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https://boardgames.stackexchange.com/questions/9446/free-city-with-babylon-b-how-to-play-the-special-ability-2nd-stage?noredirect=1 | 1,718,992,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00779.warc.gz | 121,846,252 | 36,683 | # 'Free City' with Babylon (B): How to play the special ability (2nd stage)?
You can use the special ability of Babylon (side B, 2nd stage) to get a 7th turn (EDIT: well, apparently not) where you are allowed to play your last (7th) card (which would otherwise be discarded). You can use this ability in each age (if the 2nd step is built).
How does this work in a 2-player game where the 'Free City' (NPC) uses Babylon?
Let's say the 2nd wonder stage is already built. It's the 6th turn. Player Bob manages the 'Free City'. After playing his and the NPC's card, he has left 1 card on his hand. The 'Free City' pile has 1 hidden card left. Player Alice has also 1 card left on her hand. Now the NPC gets a 7th turn.
• Bob plays his last card?
• Bob draws the last pile card and chooses which of his two cards to play?
• Bob discards his card, draws the last pile card and plays it?
• Bob transfers his card to Alice, and Alice …
• discards her card and plays the one handed over by Bob?
• discards her card, draws the last pile card and chooses which of her two cards to play?
• Alice plays her last card?
• Alice draws the last pile card and chooses which of her two cards to play?
• Alice discards her card, draws the last pile card and plays it?
(did I miss a possibility?)
Note that even if there would be no choice which card to play, it's still important which player has to play it for the 'Free City' (because this player has to decide whether to build the card or the 3rd wonder stage).
Also, if the last card of one of the players has to be played, it's important which player's hand is used, because one of the players might have the wonder Halicarnassus (which allows to build discarded cards).
• Note: The same question is asked on boardgamegeek.com.
– unor
Commented Dec 2, 2012 at 2:35
• Started formulating a read-as-written answer based on the idea that the Free City can't use special abilities (play from discard, double-play, copy guild) since you only play its action, but then there's no clear definition of whether these abilities are used as part of an action. In the end, I really don't know, but I would think it most sensible to have the player ending the round look at the last card in the draw pile and decide whether or not Babylon should play it. Commented Dec 10, 2012 at 16:44
## 1 Answer
The player controlling the Free City during the 6th turn plays the last remaining card in the Free City's draw pile.
The 7-wonders rules define each Wonder boards special abilities. (page 9)
The Hanging Gardens of Babylon the second stage gives the player the option of playing their seventh Age card instead of discarding it. This card can be played by paying its costs, discarded to earn 3 coins or used in the building of the third phase of the Wonder.
Clarification : during the sixth turn, the player can therefore play both cards they have in hand. If the second stage of the Wonder is not built, the Babylon player can then build it on their sixth turn and then play the seventh card instead of discarding it.
There isn't really a 7th turn in an Age, but it helps to think of it as an extra turn, since the resources used to build the 6th card as well as the 6th card itself are available to build the 7th and last card. This 7th card is normally the card that you would discard during the 6th turn of an Age, so that is the card that the player controlling the Free City can play to construct a structure, build their wonder, or if both if those are impossible to discard for 3 coins. (Page 7)
Clarification : The Free City cannot discard a card to get 3 coins unless it cannot play a card.
Note : during the sixth turn of the age, there is only one card remaining in the Free city’s draw pile. This last card is discarded at the same time as the seventh and last card of the players. | 923 | 3,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-26 | latest | en | 0.97899 |
https://oeis.org/A078486 | 1,632,041,856,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00345.warc.gz | 490,960,758 | 4,192 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A078486 Expansion of (x-7*x^2+19*x^3-21*x^4+10*x^5-6*x^6) / (1-9*x+31*x^2-53*x^3+44*x^4-16*x^5+6*x^6). 1
0, 1, 2, 6, 24, 102, 414, 1598, 5982, 22102, 81442, 300562, 1111638, 4117382, 15259738, 56561346, 209629750, 776850166, 2878660394, 10666717442, 39524757670, 146456879830, 542691221946, 2010931777154, 7451478924278, 27611353095414, 102313463160906 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Number of irreducible indecomposable pop-stack permutations of a certain type. LINKS Colin Barker, Table of n, a(n) for n = 0..1000 M. D. Atkinson and T. Stitt, Restricted permutations and the wreath product, Preprint, 2002. M. D. Atkinson and T. Stitt, Restricted permutations and the wreath product, Discrete Math., 259 (2002), 19-36. Index entries for linear recurrences with constant coefficients, signature (9,-31,53,-44,16,-6). MATHEMATICA CoefficientList[Series[(x-7x^2+19x^3-21x^4+10x^5-6x^6)/(1-9x+31x^2- 53x^3+ 44x^4- 16x^5+6x^6), {x, 0, 40}], x] (* Harvey P. Dale, Aug 26 2019 *) PROG (PARI) concat(0, Vec((x-7*x^2+19*x^3-21*x^4+10*x^5-6*x^6)/(1-9*x+31*x^2-53*x^3+44*x^4-16*x^5+6*x^6) + O(x^50))) \\ Colin Barker, May 27 2016 CROSSREFS Sequence in context: A141253 A306672 A324063 * A129817 A230797 A128652 Adjacent sequences: A078483 A078484 A078485 * A078487 A078488 A078489 KEYWORD nonn,easy AUTHOR N. J. A. Sloane, Jan 04 2003 EXTENSIONS Replaced definition with g.f. given by Atkinson and Stitt (2002). - N. J. A. Sloane, May 24 2016 STATUS approved
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Last modified September 19 03:31 EDT 2021. Contains 347550 sequences. (Running on oeis4.) | 757 | 2,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | latest | en | 0.588108 |
https://socratic.org/questions/544a902c581e2a79716d266f | 1,660,538,106,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572127.33/warc/CC-MAIN-20220815024523-20220815054523-00363.warc.gz | 483,264,840 | 6,015 | # Question #d266f
Oct 25, 2014
$| z - \left(2 + i\right) | < | z + 1 |$
by replacing $z$ by $x + i y$,
$R i g h t a r r o w | x + i y - \left(2 + i\right) | < | x + i y + 1 |$
by separating real parts and imaginary parts,
$R i g h t a r r o w | \left(x - 2\right) + i \left(y - 1\right) | < | \left(x + 1\right) + i y |$
by find the norms,
$R i g h t a r r o w \sqrt{{\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2}} < \sqrt{{\left(x + 1\right)}^{2} + {y}^{2}}$
by squaring,
$R i g h t a r r o w {\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2} < {\left(x + 1\right)}^{2} + {y}^{2}$
by multiplying out,
$R i g h t a r r o w {x}^{2} - 4 x + 4 + {y}^{2} - 2 y + 1 < {x}^{2} + 2 x + 1 + {y}^{2}$
by cleaning up a bit,
$R i g h t a r r o w - 2 y < 6 x - 4$
by dividing by -2,
$R i g h t a r r o w y > - 3 x + 2$
Hence, this is a region above the line $y = - 3 x + 2$.
I hope that this was helpful. | 424 | 917 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-33 | latest | en | 0.679738 |
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When trend and seasonality is present in a time series, instead of decomposing it manually to fit an ARMA model using the Box Jenkins method, another very popular method is to use the seasonal autoregressive integrated moving average (SARIMA) model which is a generalization of an ARMA model. SARIMA models are denoted SARIMA(p,d,q)(P,D,Q)[S], where S refers to the number of periods in each season, d is the degree of differencing (the number of times the…
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6 | 847 | 3,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-30 | longest | en | 0.888995 |
https://handwiki.org/wiki/Kilowatt | 1,723,735,578,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00756.warc.gz | 228,621,176 | 27,208 | # Watt
(Redirected from Kilowatt)
Short description: SI derived unit of power
watt
Unit systemSI
Unit ofpower
SymbolW
Named afterJames Watt
Conversions
1 W in ...... is equal to ...
SI base units 1 kgm2s−3
CGS units 107 ergs−1
English Engineering Units 0.7375621 ft⋅lbf/s = 0.001341022 hp
The watt (symbol: W) is the unit of power or radiant flux in the International System of Units (SI), equal to 1 joule per second or 1 kg⋅m2⋅s−3.[1][2][3] It is used to quantify the rate of energy transfer. The watt is named in honor of James Watt (1736–1819), an 18th-century Scottish inventor, mechanical engineer, and chemist who improved the Newcomen engine with his own steam engine in 1776. Watt's invention was fundamental for the Industrial Revolution.
## Overview
When an object's velocity is held constant at one meter per second against a constant opposing force of one newton, the rate at which work is done is one watt. $\displaystyle{ \mathrm{1 ~ W = 1 ~ J {/} s = 1 ~ N {\cdot} m {/} s = 1 ~ kg {\cdot} m^2 {\cdot} s^{-3}}. }$
In terms of electromagnetism, one watt is the rate at which electrical work is performed when a current of one ampere (A) flows across an electrical potential difference of one volt (V), meaning the watt is equivalent to the volt-ampere (the latter unit, however, is used for a different quantity from the real power of an electrical circuit). $\displaystyle{ \mathrm{1 ~ W = 1 ~ V \times 1 ~ A}. }$
Two additional unit conversions for watt can be found using the above equation and Ohm's law. $\displaystyle{ \mathrm{1 ~ W = 1 ~ V^2 / \Omega = 1 ~ A^2 {\cdot} \Omega}, }$ where ohm ($\displaystyle{ \Omega }$) is the SI derived unit of electrical resistance.
### Examples
• A person having a mass of 100 kg who climbs a 3-meter-high ladder in 5 seconds is doing work at a rate of about 600 watts. Mass times acceleration due to gravity times height divided by the time it takes to lift the object to the given height gives the rate of doing work or power.[lower-roman 1]
• A laborer over the course of an eight-hour day can sustain an average output of about 75 watts; higher power levels can be achieved for short intervals and by athletes.[4]
## Origin and adoption as an SI unit
The watt is named after the Scottish inventor James Watt.[5] The unit name was proposed by C. William Siemens in August 1882 in his President's Address to the Fifty-Second Congress of the British Association for the Advancement of Science.[6] Noting that units in the practical system of units were named after leading physicists, Siemens proposed that watt might be an appropriate name for a unit of power.[7] Siemens defined the unit within the existing system of practical units as "the power conveyed by a current of an Ampère through the difference of potential of a Volt".[8]
In October 1908, at the International Conference on Electric Units and Standards in London,[9] so-called international definitions were established for practical electrical units.[10] Siemens' definition was adopted as the international watt. (Also used: 1 A2 × 1 Ω.)[5] The watt was defined as equal to 107 units of power in the practical system of units.[10] The "international units" were dominant from 1909 until 1948. After the 9th General Conference on Weights and Measures in 1948, the international watt was redefined from practical units to absolute units (i.e., using only length, mass, and time). Concretely, this meant that 1 watt was defined as the quantity of energy transferred in a unit of time, namely 1 J/s. In this new definition, 1 absolute watt = 1.00019 international watts. Texts written before 1948 are likely to be using the international watt, which implies caution when comparing numerical values from this period with the post-1948 watt.[5] In 1960, the 11th General Conference on Weights and Measures adopted the absolute watt into the International System of Units (SI) as the unit of power.[11]
## Multiples
Submultiples Multiples Value SI symbol Name Value 10−1 W dW deciwatt 101 W daW decawatt 10−2 W cW centiwatt 102 W hW hectowatt 10−3 W mW milliwatt 103 W kW kilowatt 10−6 W µW microwatt 106 W MW megawatt 10−9 W nW nanowatt 109 W GW gigawatt 10−12 W pW picowatt 1012 W TW terawatt 10−15 W fW femtowatt 1015 W PW petawatt 10−18 W aW attowatt 1018 W EW exawatt 10−21 W zW zeptowatt 1021 W ZW zettawatt 10−24 W yW yoctowatt 1024 W YW yottawatt Common multiples are in bold face
Attowatt
The sound intensity in water corresponding to the international standard reference sound pressure of 1 μPa is approximately 0.65 aW/m2.[12]
Femtowatt
Powers measured in femtowatts are typically found in references to radio and radar receivers. For example, meaningful FM tuner performance figures for sensitivity, quieting and signal-to-noise require that the RF energy applied to the antenna input be specified. These input levels are often stated in dBf (decibels referenced to 1 femtowatt). This is 0.2739 microvolts across a 75-ohm load or 0.5477 microvolt across a 300-ohm load; the specification takes into account the RF input impedance of the tuner.
Picowatt
Powers measured in picowatts are typically used in reference to radio and radar receivers, acoustics and in the science of radio astronomy. One picowatt is the international standard reference value of sound power when this quantity is expressed in decibels.[13]
Nanowatt
Microwatt
Powers measured in microwatts are typically stated in medical instrumentation systems such as the electroencephalograph (EEG) and the electrocardiograph (ECG), in a wide variety of scientific and engineering instruments and also in reference to radio and radar receivers. Compact solar cells for devices such as calculators and watches are typically measured in microwatts.[14]
Milliwatt
A typical laser pointer outputs about five milliwatts of light power, whereas a typical hearing aid uses less than one milliwatt.[15] Audio signals and other electronic signal levels are often measured in dBm, referenced to one milliwatt.
Kilowatt
The kilowatt is typically used to express the output power of engines and the power of electric motors, tools, machines, and heaters. It is also a common unit used to express the electromagnetic power output of broadcast radio and television transmitters.
One kilowatt is approximately equal to 1.34 horsepower. A small electric heater with one heating element can use 1 kilowatt. The average electric power consumption of a household in the United States is about 1 kilowatt.[lower-roman 2]
A surface area of 1 square meter on Earth receives typically about one kilowatt of sunlight from the Sun (the solar irradiance) (on a clear day at midday, close to the equator).[17]
Megawatt
Many events or machines produce or sustain the conversion of energy on this scale, including large electric motors; large warships such as aircraft carriers, cruisers, and submarines; large server farms or data centers; and some scientific research equipment, such as supercolliders, and the output pulses of very large lasers. A large residential or commercial building may use several megawatts in electric power and heat. On railways, modern high-powered electric locomotives typically have a peak power output of 5 or 6 MW, while some produce much more. The Eurostar e300, for example, uses more than 12 MW, while heavy diesel-electric locomotives typically produce and use 3 and 5 MW. U.S. nuclear power plants have net summer capacities between about 500 and 1300 MW.[18]:{{{1}}}
The earliest citing of the megawatt in the Oxford English Dictionary (OED) is a reference in the 1900 Webster's International Dictionary of the English Language. The OED also states that megawatt appeared in a 28 November 1947 article in the journal Science (506:2).
Gigawatt
A gigawatt is typical average power for an industrial city of one million habitants and also the output of a large power station. The GW unit is thus used for large power plants and power grids. For example, by the end of 2010, power shortages in China's Shanxi province were expected to increase to 5–6 GW[19] and the installation capacity of wind power in Germany was 25.8 GW.[20] The largest unit (out of four) of the Belgian Doel Nuclear Power Station has a peak output of 1.04 GW.[21] HVDC converters have been built with power ratings of up to 2 GW.[22]
Terawatt
The primary energy used by humans worldwide was about 160,000 terawatt-hours in 2019, corresponding to an average continuous power consumption of 18 TW that year.[23] The most powerful lasers from the mid-1960s to the mid-1990s produced power in terawatts, but only for nanosecond intervals. The average lightning strike peaks at 1 TW, but these strikes only last for 30 microseconds.
Petawatt
A petawatt can be produced by the current generation of lasers for time scales on the order of picoseconds. One such laser is the Lawrence Livermore's Nova laser, which achieved a power output of 1.25 PW by a process called chirped pulse amplification. The duration of the pulse was roughly 0.5 ps, giving a total energy of 600 J.[24] Another example is the Laser for Fast Ignition Experiments (LFEX) at the Institute of Laser Engineering (ILE), Osaka University, which achieved a power output of 2 PW for a duration of approximately 1 ps.[25][26]
Based on the average total solar irradiance of 1.361 kW/m2,[27] the total power of sunlight striking Earth's atmosphere is estimated at 174 PW. The planet's average rate of global warming, measured as Earth's energy imbalance, reached about 0.5 PW (0.3% of incident solar power) by 2019.[28]
Yottawatt
The power output of the Sun is 382.8 YW, about 2 billion times the power estimated to reach Earth's atmosphere.[29]
## Conventions in the electric power industry
In the electric power industry, megawatt electrical (MWe[30] or MWe)[31] refers by convention to the electric power produced by a generator, while megawatt thermal or thermal megawatt[32] (MWt, MWt, or MWth, MWth) refers to thermal power produced by the plant. For example, the Embalse nuclear power plant in Argentina uses a fission reactor to generate 2109 MWt (i.e. heat), which creates steam to drive a turbine, which generates 648 MWe (i.e. electricity). Other SI prefixes are sometimes used, for example gigawatt electrical (GWe). The International Bureau of Weights and Measures, which maintains the SI-standard, states that further information about a quantity should not be attached to the unit symbol but instead to the quantity symbol (e.g., Pth = 270 W rather than P = 270 Wth) and so these unit symbols are non-SI.[33] In compliance with SI, the energy company Ørsted A/S uses the unit megawatt for produced electrical power and the equivalent unit megajoule per second for delivered heating power in a combined heat and power station such as Avedøre Power Station.[34]
When describing alternating current (AC) electricity, another distinction is made between the watt and the volt-ampere. While these units are equivalent for simple resistive circuits, they differ when loads exhibit electrical reactance.
Radio stations usually report the power of their transmitters in units of watts, referring to the effective radiated power. This refers to the power that a half-wave dipole antenna would need to radiate to match the intensity of the transmitter's main lobe.
## Distinction between watts and watt-hours
The terms power and energy are closely related but distinct physical quantities. Power is the rate at which energy is generated or consumed and hence is measured in units (e.g. watts) that represent energy per unit time.
For example, when a light bulb with a power rating of 100W is turned on for one hour, the energy used is 100 watt hours (W·h), 0.1 kilowatt hour, or 360 kJ. This same amount of energy would light a 40-watt bulb for 2.5 hours, or a 50-watt bulb for 2 hours.
Power stations are rated using units of power, typically megawatts or gigawatts (for example, the Three Gorges Dam in China, is rated at approximately 22 gigawatts). This reflects the maximum power output it can achieve at any point in time. A power station's annual energy output, however, would be recorded using units of energy (not power), typically gigawatt hours. Major energy production or consumption is often expressed as terawatt hours for a given period; often a calendar year or financial year. One terawatt hour of energy is equal to a sustained power delivery of one terawatt for one hour, or approximately 114 megawatts for a period of one year:
Power output = energy / time
1 terawatt hour per year = 1×1012 W·h / (365 days × 24 hours per day) ≈ 114 million watts,
equivalent to approximately 114 megawatts of constant power output.
The watt-second is a unit of energy, equal to the joule. One kilowatt hour is 3,600,000 watt seconds.
While a watt per hour is a unit of rate of change of power with time,[lower-roman 3] it is not correct to refer to a watt (or watt-hour) as a watt per hour.[35]
## Explanatory notes
1. The energy in climbing the stairs is given by mgh. Setting m = 100 kg, g = 9.8 m/s2 and h = 3 m gives 2940 J. Dividing this by the time taken (5 s) gives a power of 588 W.
2. Average household electric power consumption is 1.19 kW in the US, 0.53 kW in the UK. In India it is 0.13 kW (urban) and 0.03 kW (rural) – computed from GJ figures quoted by Nakagami, Murakoshi and Iwafune.[16]
3. Watts per hour refers to the rate of change of power being used (or generated). For example, a power plant that changes its power output from 100 MW to 200 MW in 15 minutes would have a ramp-up rate of 400 MW/h. Gigawatts per hour are used to characterize the ramp-up required of the power plants on an electric grid to compensate for loss of output from other sources, such as when solar power generation drops to zero as the sun sets. See duck curve.
## References
1. Newell, David B; Tiesinga, Eite (2019). The international system of units (SI) (Report). Gaithersburg, MD: National Institute of Standards and Technology. doi:10.6028/nist.sp.330-2019. §2.3.4, Table 4.
2. Yildiz, I.; Liu, Y. (2018). "Energy units, conversions, and dimensional analysis". in Dincer, I.. Comprehensive energy systems. Vol 1: Energy fundamentals. Elsevier. pp. 12–13. ISBN 9780128149256.
3. Avallone, Eugene A, ed. (2007), Marks' Standard Handbook for Mechanical Engineers (11th ed.), New York: Mc-Graw Hill, pp. 9–4, ISBN 978-0-07-142867-5 .
4. Klein, Herbert Arthur (1988). The Science of measurement: A historical survey. New York: Dover. p. 239. ISBN 9780486144979.
5. "Address by C. William Siemens". Report of the Fifty-Second meeting of the British Association for the Advancement of Science. 52. London: John Murray. 1883. pp. 1–33.
6. Siemens supported his proposal by asserting that Watt was the first who "had a clear physical conception of power, and gave a rational method for measuring it." "Siemens, 1883, p. 6"
7. "Siemens", 1883, p. 5"
8. Tunbridge, P. (1992). Lord Kelvin: His Influence on Electrical Measurements and Units. Peter Peregrinus: London. p. 51. ISBN 0-86341-237-8.
9. Fleming, John Ambrose (1911). "Units, Physical". in Chisholm, Hugh. Encyclopædia Britannica. 27 (11th ed.). Cambridge University Press. pp. 738–745; see page 742.
10. "Resolution 12 of the 11th CGPM (1960)". Bureau International des Poids et Mesures (BIPM).
11. Ainslie, M. A. (2015). A century of sonar: Planetary oceanography, underwater noise monitoring, and the terminology of underwater sound. Acoustics Today.
12. Morfey, C.L. (2001). Dictionary of Acoustics.
13. "Bye-Bye Batteries: Radio Waves as a Low-Power Source", The New York Times, Jul 18, 2010 .
14. Stetzler, Trudy; Magotra, Neeraj; Gelabert, Pedro; Kasthuri, Preethi; Bangalore, Sridevi. "Low-Power Real-Time Programmable DSP Development Platform for Digital Hearing Aids". Datasheet Archive.
15. Nakagami, Hidetoshi; Murakoshi, Chiharu; Iwafune, Yumiko (2008). "International Comparison of Household Energy Consumption and Its Indicator". ACEEE Summer Study on Energy Efficiency in Buildings. Pacific Grove, California: American Council for an Energy-Efficient Economy. Figure 3. Energy Consumption per Household by Fuel Type. 8:214–8:224. Retrieved 14 February 2013.
16. Elena Papadopoulou, Photovoltaic Industrial Systems: An Environmental Approach, Springer 2011 ISBN:3642163017, p.153
17. (in en-us) 2007–2008 Information Digest (Report). 19. United States Nuclear Regulatory Commission. 2007-08-01. pp. 84-101.
18. Bai, Jim; Chen, Aizhu (11 November 2010). "China's Shanxi to face 5–6 GW power shortage by yr-end – paper". in Lewis, Chris. Reuters.
19. "Not on my beach, please". The Economist. 19 August 2010.
20. "Chiffres clés" (in fr). Electrabel. 2011.
21. Davidson, CC; Preedy, RM; Cao, J; Zhou, C; Fu, J (October 2010), "Ultra-High-Power Thyristor Valves for HVDC in Developing Countries", 9th International Conference on AC/DC Power Transmission, London: IET .
22. Hannah Ritchie; Max Roser (2020). "Global Direct Primary Energy Consumption". Our World in Data (Published online at OurWorldInData.org.). Retrieved 2020-02-09.
23. "Crossing the Petawatt threshold". Livermore, California: Lawrence Livermore National Laboratory.
24. Loeb, Norman G.; Johnson, Gregory C.; Thorsen, Tyler J.; Lyman, John M. et al. (15 June 2021). "Satellite and Ocean Data Reveal Marked Increase in Earth's Heating Rate". Geophysical Research Letters 48 (13). doi:10.1029/2021GL093047. Bibcode2021GeoRL..4893047L.
25. Williams, David R.. "Sun Fact Sheet". NASA.
26. Cleveland, CJ (2007). "Watt". Encyclopedia of Earth.
27. "Solar Energy Grew at a Record Pace in 2008 (excerpt from EERE Network News". Department of Energy). 25 March 2009.
28. "Inverter Selection". Northern Arizona Wind and Sun. | 4,643 | 17,725 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-33 | latest | en | 0.850935 |
https://mathematica.stackexchange.com/questions/172275/checking-rationality-using-findinstance | 1,722,962,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640497907.29/warc/CC-MAIN-20240806161854-20240806191854-00197.warc.gz | 302,477,560 | 42,284 | # Checking Rationality Using FindInstance
I need to find example if $x,\sqrt{x^2+6},\sqrt{x^2+12}$ all can be rational at once or not. But the following command FindInstance[Element[Sqrt[x + 6], Rationals], {x}]returns an errors saying
The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist.
Anyway to bypass this or modify the input to get result?
Edit FindInstance[Element[y - 8.4, Rationals], {y}] also returns the same error. Is this a bug or what?
• Please, elaborate what means "all can be rational at once or not" Commented Apr 30, 2018 at 17:53
• @JoséAntonioDíazNavas for example both √y and √(y+9) are rational for y =16. I need to x such that all three of those terms are rational Commented Apr 30, 2018 at 18:35
Maybe this?:
FindInstance[6 + x^2 == y^2 && 12 + x^2 == z^2, {x, y, z}, Rationals]
{{x -> -(1/2), y -> -(5/2), z -> 7/2}}
Given that by definition $|c|=\sqrt{c^2}$ for $c\in\mathbb{R}$, then the solutions should be:
{{x -> -(1/2), y -> 5/2, z -> 7/2}} || {{x -> 1/2, y -> 5/2, z -> 7/2}}
You might try brute-force search, enumerating all (positive) rationals starting from 0:
x = 0; While[! (Element[Sqrt[x^2 + 6], Rationals] &&Element[Sqrt[x^2 + 12], Rationals]), x = 1/(Floor[x] + 1 - FractionalPart[x])]; x
with simple answer: (* x = 1/2 *).
• relying on a bit of luck that the answer occurs early in the sequence. This could take approximately forever. Commented May 2, 2018 at 19:29
• Above one-liner check 10^6 distinct rationals/second. You can easily get 10^8/sec rewriting in C. Commented May 3, 2018 at 18:41
• Ah, but the space of "all" rationals is infinite. It is a fair point that such a puzzle likely has a "nice" answer though. Commented May 3, 2018 at 19:07
• Space of all rationals representable in computer is finite, e.g. limited by memory. FindInstance probably uses algorithm like this for Integers, so why not for Rationals? Commented May 4, 2018 at 5:27
A systematic look-up for x^2 == y && 9 + y == z^2
sol = Union[{x1/x2, x1^2/x2^2, z1/z2} /.
Solve[(9 + (x1/x2)^2 == (z1/z2)^2 && 0 <= x1 <= a && 0 < x2 <= a &&
0 <= z1 <= a && 0 < z2 <= a) /. a -> 100, {x1, x2, z1, z2},
Integers, Method -> Reduce]]
(* {{0, 0, 3}, {13/28, 169/784, 85/28}, {11/20, 121/400, 61/20},
{16/21, 256/441, 65/21}, {7/8, 49/64, 25/8}, {5/4, 25/16, 13/4},
{8/5, 64/25, 17/5}, {28/15, 784/225, 53/15}, {9/4, 81/16, 15/4},
{65/24, 4225/576, 97/24}, {20/7, 400/49, 29/7}, {63/20, 3969/400, 87/20},
{55/16, 3025/256, 73/16}, {4, 16, 5}, {56/11, 3136/121, 65/11},
{45/8, 2025/64, 51/8}, {80/13, 6400/169, 89/13}, {77/12, 5929/144, 85/12},
{36/5, 1296/25, 39/5}, {35/4, 1225/16, 37/4}, {72/7, 5184/49, 75/7},
{40/3, 1600/9, 41/3}} *)
and for 6 + x^2 == y^2 && 12 + x^2 == z^2
sol2 = Union[{x1/x2, y1/y2, z1/z2} /.
Solve[(6 + (x1/x2)^2 == (y1/y2)^2 && 12 + (x1/x2)^2 == (z1/z2)^2 &&
0 <= x1 <= a && 0 < x2 <= a && 0 <= y1 <= a && 0 < y2 <= a &&
0 <= z1 <= a && 0 < z2 <= a) /. a -> 100, {x1, x2, y1, y2, z1,
z2}, Integers, Method -> Reduce]]
(* {{1/2, 5/2, 7/2}} *)
Seems to the the only solution. | 1,284 | 3,105 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-33 | latest | en | 0.864373 |
http://mathcrush.com/math_art_worksheets.html | 1,506,417,105,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695375.98/warc/CC-MAIN-20170926085159-20170926105159-00107.warc.gz | 206,808,144 | 17,917 | # Math Art - Worksheets
Scroll down to see all choices.
## Ordinal Art - Level 1
These one page worksheets introduce ordinals. Example: 1st, 2nd, 3rd, etc. Students color certain objects to recognize their place in a set. There is also a spelling section on each handout.
## Word and Number Form - Art Level 1
This one page, art worksheet reviews word and number forms. Students read the word form of a number and then find and color the flower that matches each answer.
Key concept: Students need to learn how to spell numbers and see both the number and word form of numbers.
## Answer, Find, and ColorPlace ValueAll Levels
This CHRISTMAS art worksheet reviews place value. Students write the correct number for each place value and then color the object with the corresponding letter.
## Greater Than and Less ThanLevel 1
This one page worksheet introduces greater than and less than. It uses alligators to help explain it. Students cut out alligator heads and glue the faces to show which number is bigger.
Key concept: Understanding inequalities and their symbols (<, >).
This art worksheet has students write numbers in standard form and includes a few place value questions.
Key concept: Students should be able to count, read, and write whole numbers up to 100,000, and identify the place value of specific digits in numbers.
## Basic Subtraction - Level 1
This two page worksheet introduces subtraction. The first page is basic practice and the second page has two Answer, Find, and Color activities.
Important note: These handouts were designed to go with a short basic subtraction video which is currently located on YouTube, but can also be used alone.
This art worksheet reviews USA Currency. Students count dollars and change to calculate an amount.
Key concept: Students should be able to solve problems using different combinations of coins and bills.
## Stained GlassMoney Review - Level 1
This art worksheet reviews money (USA Currency). It includes basic change, rounding to the nearest dollar, and adding and subtracting money. Each student chooses their own colors to create a unique art project.
Key concept: Understand money amounts in decimal notation.
This art worksheet reviews time. Students read clocks, count minutes, and estimate hours.
Key concept: Students should be able to tell time, and determine the duration of intervals of time in minutes and hours.
## Time Art - Level 2
This one page, art worksheet helps students practice time. Students answer the word problems, find the answers in the grid, and then shade the squares that match the answers.
These one page art worksheets have students practice reading a schedule. The schedule and picture become more difficult as the levels get higher. For example: Information is left out of the schedule and students must calculate what is missing.
These one page art worksheets review multi-digit addition. Students answer the problems, find the answers at the bottom of the page, and shade in the corresponding number in the picture. The picture is symmetric.
## Multiple Digit Subtraction - Art Level 3
This one page, art worksheet reviews multiple digit subtraction with borrowing. Students answer the problems, and then find and shade in the square that matches each answer.
Helpful idea: Students can use different colors to create more interesting balloons.
These one page art worksheets review multi-digit subtraction. Students answer the problems, and color in the corresponding number for each answer. Odd answers will be one color and even answers another. Questions are different but the picture is the same.
## Add and Subtract ArtLevel 2
This one page worksheet covers addition and subtraction. Students answer the problems, and then find and color the cat that matches each answer. It includes borrowing and carrying.
This art worksheet has students use their prior knowledge of addition to get a better understanding of multiplication. It includes multiple choice questions, grouping, and number lines.
Key concept: Use repeated addition, grouping, and counting by multiples to do multiplication.
This art worksheet reviews basic multiplication. It includes solving multiplication with grouping, basic algebra, money, and a few word problems.
Key concept: Practice makes perfect, especially with multiplication.
## The Basics Art - Level 1
This one page worksheet covers addition, subtraction, multiplication, and division. Students choose two colors and answer the questions. After they answer the problems they color the corresponding shapes to create a picture. This worksheet can be used as a HALLOWEEN handout.
## Fill in the Blanks ArtLevel 2
This one page HALLOWEEN worksheet is an addition, subtraction, multiplication, and division practice page. It uses fill in the blanks to help students review. Students answer the problems, and then find and color the object that matches each answer.
These one page art worksheets review multi-digit multiplication. The problems are multiple choice to help with test taking skills. The problems are different but the picture is the same for each level.
These one page CHRISTMAS worksheets review long division. The drawing is the same but the questions are different for each level.
These one page art worksheets review multi-digit division. Students should already know long division without remainders or decimals. They answer the problems, find the answers at the bottom of the page, and shade in the corresponding number in the picture. Questions different but the picture the same.
## Multiplying and Dividing by Multiples of 10Level 2
This one page worksheet reviews multiplying and dividing by multiples of 10. It uses simple multiples (10, 100, and 1,000), and also includes a few problems with slightly more difficult ones (300 and 400).
Key concept: Students should know basic multiplication and division using multiples of ten which will help them develop better strategies for multiplication and division with larger numbers, and estimating.
## Help Video
Vimeo: https://vimeo.com/171025484
Links for you to copy and paste for students
This math art page uses color shading to create a picture. Students divide by multiples of ten. This is great practice for estimating two digit division.
Helpful idea: Use worksheet for special occasion like Mother's Day, birthday, or thank you note. Students can fill in the empty banner.
This math art page uses shading to create a picture. Students use long division to solve the problems.
Student misunderstanding: Most students get lost with two digit division. Estimate the divisor (number on the outside) to a multiple of ten, and use that number to help find the quotient (answer). Use Division Book - Level 2 pages 38 to 47 for review.
These one page, art worksheets review the four basic operations of arithmetic (addition, subtraction, multiplication, and division). The drawing is the same but the questions and answers are different for each level.
## Stained GlassRounding Review - Level 1
This art worksheet reviews rounding. Students round to the nearest ten, hundred, and thousand. Each student chooses their own colors to create a unique art project.
Helpful idea: Students should be able to color or shade in shapes neatly, but if students are struggling with the art use it as extra credit to motivate practice.
This one-page art worksheet has two pictures hidden inside. Students will need to round each number to a specific value. They must answer all the questions but choose between shading in the odd or even numbered problems to generate one of the two different pictures hidden in the worksheet.
## Estimating - Level 1
This one page HALLOWEEN worksheet is an estimation practice page. Students estimate the problems to the nearest whole number. After they answer the problems they shade in the corresponding shapes to create a scary picture.
Key concept: Students need to understand the difference between whole numbers and decimals (pieces).
This free art worksheet reviews basic arithmetic (add, subtract, multiply, and divide) and estimation.
Student misunderstanding: Students should estimate before doing any operations. Example: 13+29 should be estimated to 10+30=40, NOT 13+29=42 and then estimated to 40.
These one page art worksheets review adding with decimals. Students will need to line up the decimals and digits and add.
Helpful idea: Use picture as extra credit.
These one page art worksheets review subtracting with decimals. Students will need to line up the decimals and digits and subtract.
Key concept: Subtract with decimals.
These one page worksheets cover adding and subtracting decimals. Each worksheet includes basic practice and an Answer, Find, and Shade Activity.
Important note: These handouts are designed to go with a short Adding and Subtracting Decimals Video which is currently located on YouTube, but can also be used alone.
These one page, art worksheets review multiplying decimals. The drawing is the same but the questions are different for each level.
Student misunderstanding: Count how many digits are to the right of the decimal in all the numbers before multiplying. After the final answer is calculated put the decimal back so the same number of digits are to the right.
This art worksheet reviews the basic operations (add, subtract, multiply, and divide) with decimals.
Student misunderstanding: Multiplying: the total number of digits to the right of the decimal should be the same in the final answer. Dividing: cannot have a decimal in the divisor (outside number). Whatever you do to the outside you have to do to the inside.
This one page art worksheet reviews basic decimals. It includes changing simple fractions to decimals, understanding the word form, plotting decimals on a number line, and two word problems.
Student misunderstanding: Students should understand the difference between tenths and hundredths (Example: 0.4 and 0.04.) and know that 0.4 and 0.40 are equal.
## Answer, Find, and ColorDecimal Review - Level 2
This art worksheet reviews decimals. It covers word form of decimals, greatest decimal value, rounding, and problem solving. Students choose two different colors and color the odd numbered problems one color and the even ones the other color.
Key concept: Students understand decimals and solve problems involving decimals.
This math art page uses shading to create a picture. Students find the common denominator to add fractions.
Student misunderstanding: Students need to know their basic multiples to find the common denominator. It's a good idea to practice basic multiplication facts on a regular basis.
## Answer, Find, and ShadeSubtracting Fractions with Unlike Denominators - Level 2
This math art page uses shading to create a picture. Students find the common denominator to subtract fractions.
Student misunderstanding: Students need to know their basic multiples to find the common denominator. It's a good idea to practice basic multiplication facts on a regular basis.
This math art page uses shading to create a picture. Students divide fractions using the reciprocal method and cross canceling to simplify.
Student misunderstanding: When dividing fractions change the division to multiplication and flip the fraction on the right.
## Mixed Problems with FractionsLevel 1
This one-page art worksheet covers arithmetic with fractions. Students use their knowledge of basic operations to solve fraction computations.
Key concept: Students perform calculations and solve problems involving addition, subtraction, and simple multiplication and division of fractions.
## Stained GlassEstimating Fractions - Level 2
This art worksheet covers estimating fractions. Students need to estimate the mixed numbers to the nearest whole number and then solve. Each student chooses their own colors to create a unique art project.
Student misunderstanding: Estimating is something to make your life easier, so students need to estimate before trying to solve the problem.
## Stained Glass - Level 2Comparing Fractions and Decimals
This one page art worksheet covers comparing fractions and decimals. Students circle the greatest amount in each group and then color the related shapes in the picture. Each student chooses their own colors to create a unique art project.
Key concept: Students need to be able to compare fractions and decimals and be able to visualize the size of a number.
This math art page uses shading to create a picture. Students use their factoring skills to find the GCF.
Key concept: Factoring is the key to simplifying fractions. If a student can factor / divide then they usually understand simplifying.
## Mean, Median, Mode, RangeLevel 2
This one page HOLIDAY art worksheet reviews mean, median, mode, and range. Students answer the problems, and then find and color the turkey that matches each answer.
Key concept: Students should know what each term means and how to find the solution.
## Answer, Find, and Shade (2 in 1)Mean, Median, Mode, and RangeLevel 2
This one-page art worksheet has two pictures hidden inside. Students will need to find the mean, median, mode, and range for each set. They must answer all the questions but get to choose to only shade in the odd numbered or even numbered problems to get one of the two different pictures.
## Stained GlassUnit Rate - Level 2
We call this art worksheet Stained Glass. The picture is symmetrical and students choose their own colors to create a unique art project. The worksheet covers unit rate and students answer the problems, find the numbers in the picture, and color the shapes based on the colors they chose.
Key concept: Understand that rate is a measure of one quantity per unit value of another quantity.
These one page art worksheets review adding integers. Students are given two or three numbers to add. Each level has different questions, but the picture and answers are the same.
These one page art worksheets review subtracting integers. Students are given two or three numbers to subtract. Each level has different questions, but the picture and answers are the same.
These one page abstract art worksheets have two pictures hidden inside. Students are given two or three numbers to multiply or divide. They must answer all the questions but get to choose to only shade in the odd numbered or even numbered problems to get one of the two different pictures. They will be more motivated to finish because they will want to compare their work.
## IntegersMultiplying and Dividing Art
These one page, art worksheets cover multiplying and dividing integers. Students pick their own colors and answer each problem. Then they find and color the character that matches each answer.
Helpful idea: Even if students have not learned about integers or order of operations you can use it as a quick introduction and then review their basics.
This one page art worksheet reviews arithmetic with rational numbers, positive and negative integers, fractions, and decimals.
Key concept: Students need to know and review their basics to progress through higher math.
An alternative version of Dividing by 10's (above). Students use the Order of Operations to solve expressions.
Teacher misunderstanding: Many teachers like to use the phrase, "Please Excuse My Dear Aunt Sally," to help students remember the Order of Operations. This is helpful, but many students then think that multiplication comes before division and addition comes before subtraction.
This one page art HALLOWEEN worksheet reviews integers and order of operations.
Key concept: Understanding positive and negative numbers and the order of operations.
These one page Halloween worksheets review verbal sentences and variables. Students read the sentences (equations) and find the missing number. The drawing is the same but the questions are different for each level.
These one page, art worksheets review the distributive property. Level one has students fill in the blanks, and level 2 and 3 has them solve equations. The drawing is the same but the questions are different for each level.
These one page worksheets have students use the distributive, associative, and commutative properties to simplify expressions. The drawing is the same but the shading and questions are different for each worksheet.
Version 1
Version 2
This one page art worksheet reviews squares and square roots. The square roots are perfect squares so students will not have to estimate and no irrational numbers will be left over. Watch out for the decimals.
## Answer, Find, and Color - Level 1Fill in the Blank
This free winter holiday, one page art worksheet reviews basic order of operations by having students fill in the blanks to complete various equations.
Help idea: Allow students to choose their own colors so the art work becomes more personal and unique. Just shade in lightly.
These one, page art worksheets use fill in the blank questions to review order of operations and solving equations. The drawing is the same but the questions are different for each level.
Helpful idea: Use the picture to help students see patterns. If they do the worksheet incorrectly than the shape will not look correct.
This one page art worksheet reviews substitution, order of operations, and basic algebra. Students substitute the values into the expressions and solve.
Student misunderstanding:
1. Variables next to each other or a number next to
a variable is the same as multiplication.
For Example: abc = a x b x c or 2h = 2 x h
This one page art worksheet reviews substitution, order of operations, and basic algebra. Students substitute the values into the expressions and solve.
Student misunderstanding:
1. Variables next to each other mean multiplication.
For Example: abc = a x b x c
2. Be careful replacing the decimal after multiplying.
For Example: 0.6 x 0.6 = 0.36, but 0.6 x 6 = 3.6
This one page art worksheet reviews substitution, order of operations, and basic algebra. Students substitute the values into the expressions and solve.
Student misunderstanding: Variables next to each other mean multiplication.
For Example: abc = a x b x c.
These one page art worksheets review solving equations, substitution, charts, and learning real world equations. Students are given a chart and must fill in the empty spaces. Level one and two include equations to help students, but the level 3 worksheet makes students research or use prior knowledge.
## Solving One Step EquationsLevel 2
This one page HALLOWEEN worksheet is a basic algebra practice page. Students use addition, subtraction, multiplication, and division to solve one step equations. After they answer the problems they shade in the corresponding shapes to create a scary picture.
Key concept: Solving variables and basic equations.
## One-Step EquationsLevel 2
This one page WINTER HOLIDAY worksheet covers solving one-step equations. It includes addition, subtraction, multiplication, and division. Students answer the problems and then shade in the corresponding shapes to create a picture.
## Combining Like Terms ArtLevel 1
This one page, art worksheet has students practice combining like terms. It includes using the Distributive Property to simplify the expressions.
Student misunderstanding: Even though a part of an expression or equation is surrounded by parenthesis, it can still be pulled apart using the Distributive Property. For example: 3(y + 4) = 3y + 12.
## Two-Step Fill in the Blanks
These one page art worksheets cover solving two-step equations. They help students visualize the process which should lead to solving equations faster.
Helpful idea: If students are confused change the blanks to a variable and solve the equation.
## Answer, Find, and ColorMulti-Step Equations
These one page art worksheets review multi-step equations. Level one includes combining like terms, Level two includes The Distributive Property, and Level three includes fractions.
This math art worksheet reviews solving equations. Students answer the problems, and then find and shade in the area that matches each answer.
Key concept: Understanding that we can change equations or "move numbers and variables" as long as we do the same to both sides.
These one page Scary Halloween art worksheets review solving functions. Students need to substitute the given number into the function and solve. The picture and answers are the same but the problems are different for each level.
This one page, art worksheet reviews solving inequalities. Students solve two inequalities and find the answer they have in common.
Student misunderstanding: Multiplying or dividing each side of an inequality by a negative number reverses the inequality sign. To prove this have students check their answers.
This art worksheet covers decimals and percents. Students look at squares and write the decimal and percent of the shaded part. In addition, there are a few find the percent of a number problems.
Key concepts: Interpret percents as a part of a hundred, find decimal and percent equivalents for common fractions, and compute a given percent of a number.
These one page art worksheets have two pictures hidden inside. Students need to use their knowledge of percent of numbers to solve various equations. Examples: 42.5% of 59 = ?, ? % of 85 = 10, or 40% of ? = 81. They must answer all the questions but for the shading students have the option to only shade in the odd numbered or even numbered problems to get one of the two different pictures.
These one page, art worksheets review percent of a number. Students need to find the percent of a number, what percent one number is of another, and find the original number when a percent is given.
Helpful idea: The two versions have the same questions, but in slightly different order and the pictures have been altered. Give half the class one version and the rest the other.
These one page art worksheets review percent of change. Students are given an original and new number and must figure out if the change is an increase or decrease and then calculate the percent of change. After solving, students find the answers in the picture and shade in the corresponding shapes.
These one page art worksheets review fractions, decimals, and percent. Problems include finding the greatest number, percent and fraction of a number, arithmetic with fractions and decimals, and a word problem.
This FREE one page art worksheet reviews Probability. It is a tribute to the 2014 FIFA World Cup. Students read word problems and determine the probability of a certain event occurring.
Key concept: Students make predictions for simple probability situations.
These one page, art worksheets review proportions. Students find the missing value in each proportion. The drawing is the same but the questions are different for each level.
Helpful idea: The pattern is seen best from a distance.
This art worksheet covers linear measurement. Students find the distance of line segments using number lines and equations.
Key concept: Students should be able to calculate the distance between two points and read, understand, and visualize numbers on a number line.
This art worksheet includes word problems on distance. Students use a distant chart in kilometers to answer various questions, and convert kilometers to miles or miles to kilometers.
Important note: To shade the picture correctly students must use 1 kilometer = 0.62 miles when converting.
This one page, art worksheet reviews finding the slope from two points. Students calculate the slope using the change in x and y.
Helpful idea: When subtracting the coordinates, change the second point to its opposite.
Example: Given two points: (3 , 4);(2 , -2) - subtract
( 3 , 4)
(-2 , +2) opposite of the numbers
( 1 , 6) , answer: slope is 6.
## Word Problems Art - All Levels
These one page WINTER HOLIDAY worksheets are simple words problems designed for all levels. Note: The picture and answers are the same for each level, but the questions are different.
Level 1 Print
Level 2 Print
Level 3 Print
These one page, art worksheets review word problems. The drawing is the same but the questions are different for each level.
Can you see it?
This one-page art worksheet covers vertical and adjacent angles. Students use their knowledge of lines and vertical angles to calculate missing angles.
Key concept: Use facts about supplementary, complementary, vertical, and adjacent angles to calculate unknown angles in a figure.
## Graphing / Coordinate Plane - A
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Which number is the x-axis and which one is the y-axis? Have to go into the elevator before you can go up or down, or have to move the ladder before you climb it. So the first number is the x-axis (number going across).
## Graphing / Coordinate Plane - B
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Where is the origin? The origin is where the x-axis and y-axis cross. It is also where the zero goes for both axes.
## Graphing / Coordinate Plane - C
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Which number is the x-axis and which one is the y-axis? Have to go into the elevator before you can go up or down, or have to move the ladder before you climb it. So the first number is the x-axis (number going across).
## Graphing / Coordinate Plane - D
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Where is the origin? The origin is where the x-axis and y-axis cross. It is also where the zero goes for both axes.
## Graphing / Coordinate Plane - E
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Which number is the x-axis and which one is the y-axis? Have to go into the elevator before you can go up or down, or have to move the ladder before you climb it. So the first number is the x-axis (number going across).
## Graphing / Coordinate Plane - FLevel 3
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Which number is the x-axis and which one is the y-axis? Have to go into the elevator before you can go up or down, or have to move the ladder before you climb it. So the first number is the x-axis (number going across).
## Line of Symmetry Art- All Levels
These one page Easter worksheets use line of symmetry and art to review student knowledge. Students re-draw the second half of the shape to create the full rabbit.
Key concept: Line of symmetry is a line such that if a figure is folded about the line, then one half of the figure matches the other half.
## Angle Art - Level 2
Students need to find the missing angle for each polygon. This one page worksheet covers triangles, quadrilaterals, pentagons, and hexagons. Students choose two colors and answer the questions. After they answer the problems they color the corresponding shapes to create a picture. This worksheet can be used as a HALLOWEEN handout.
Alternative version to The Basics Art - Level 1.
## Answer, Find, and Shade - Level 3Distance of a Line Segment
This one page art worksheet reviews finding the distance of a line segment. Students will need to know the Distance Formula, but also can use the Pythagorean Theorem on most of the problems. The problems include line segments on graphs and two coordinates with no graph.
This one page art worksheet reviews the distance formula. Students find the distance between two points. All answers are rational numbers. This worksheet can also be used during Halloween.
Helpful idea: Students should be allowed to use a calculator.
These one page worksheets cover converting capacity. There is one worksheet for the USA Customary System of Measurement and one for the Metric System. The drawing is the same but the questions and answers are different.
USA Customary System of Measurement
Metric System | 5,631 | 28,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-39 | latest | en | 0.912684 |
http://docs.itascacg.com/3dec700/common/models/cwipp/doc/modelcwipp.html | 1,670,591,177,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00265.warc.gz | 11,378,138 | 7,396 | WIPP-Salt Model
A crushed-salt constitutive model is implemented in FLAC3D/3DEC to simulate volumetric and deviatoric creep compaction behaviors. The model is a variation of the WIPP model, and is based on the model described by Sjaardema and Krieg (1987), with an added deviatoric component as proposed by Callahan and DeVries (1991).
Formulations
In the crushed-salt constitutive model, the material density, $$\rho$$, is a variable that evolves as a function of compressive volumetric strain, $$\epsilon_v$$, from the initial crushed-salt emplacement value, $${\rho}_i$$, to the ultimate intact salt density, $${\rho}_f$$. The relation between the rate of change of volumetric strain and density for use in the FLAC3D/3DEC incremental Lagrangian formulation may be outlined as follows. (Remember that, as a convention, stresses and strains are negative in compression.)
Consider a given material domain of mass, $$m$$ (which at time, $$t$$, has volume, $$V_{\circ}$$, and density, $${\rho}_{\circ}$$), and let the volumetric strain increment, $$\Delta \epsilon_v$$, correspond to a change in volume, $$\Delta V$$, and in density from $$\rho_{\circ}$$ to $$\rho$$ during the time interval, $$\Delta t$$. By virtue of mass conservation, we have
(1)$\rho_{\circ} V_{\circ} = \rho (V_{\circ} + \Delta V)$
and, by definition of volumetric strain, we obtain
(2)$\rho = {{\rho_{\circ}}\over{1 + \Delta \epsilon_v}}$
Also, from a continuum approach, we may write
(3)$\rho = {{m}\over{V}}$
and the rate-of-change of density of the given mass is
(4)$\dot{\rho} = -{{m}\over{V^2}} \dot{V}$
Using $$\dot{\epsilon_v} = \dot{V} / V$$ together with Equation (3), after some manipulation we obtain
(5)$\dot{\epsilon_v} = -{{\dot{\rho}}\over{\rho}}$
A measure of the crushed-salt compaction is given by the fractional density, $$F_d$$, defined as the ratio between actual and ultimate salt densities:
(6)$F_d = {{\rho}\over{\rho_f}}$
In the model implementation, it is assumed that the creep-compaction mechanism is irreversible (the density can only increase and cannot decrease) and bounded (no further compaction occurs after the intact salt value has been reached).
Constitutive Equations
In the crushed-salt model, elastic stress and strain rates are related by means of the incremental expression of Hooke’s law:
(7)$\dot{\sigma}_{ij} = 2G \left[\dot{\epsilon}_{ij}^e -{{\dot{\epsilon}_{kk}^e}\over{3}} \delta_{ij} \right] + K \dot{\epsilon}_{kk}^e \delta_{ij}$
where $${\delta}_{ij}$$ is the Kronecker delta.
In this expression, the bulk modulus, $$K$$, and shear modulus, $$G$$, are related to the density by a nonlinear empirical law of the form
(8)$K = K_f e^{K_1(\rho - \rho_f)}$
(9)$G = G_f e^{G_1(\rho - \rho_f)}$
where $$\rho_f$$, $$K_f$$, and $$G_f$$ are properties of the intact salt, and $$K_1$$, $$G_1$$ are two constants determined from the condition that bulk and shear must take their initial values at the initial value of the density.
It is assumed that, for density values below that of the intact salt, the total strain-rate $$\dot\epsilon_{ij}$$ can be expressed as the sum of three contributions: nonlinear elastic, $$\dot\epsilon_{ij}^e$$; viscous compaction, $$\dot\epsilon_{ij}^c$$; and viscous shear, $$\dot\epsilon_{ij}^v$$. The elastic strain-rate takes the form
(10)$\dot\epsilon_{ij}^e = \dot\epsilon_{ij} - \dot\epsilon_{ij}^c - \dot\epsilon_{ij}^v$
The viscous compaction term is based on an experimental compaction-rate law of the form
(11)$\dot{\rho}^c = -B_0 \left[ 1 - e^{-B_1 \sigma} \right] e^{B_2 \rho}$
where $$\sigma = \sigma_{kk} / 3$$ is the mean stress, and $$B_0$$, $$B_1$$, $$B_2$$ are constants determined experimentally from results of isotropic compaction tests.
The volumetric compaction strain-rate $$\dot\epsilon_v^c$$ may be derived after substitution of Equation (11) for $$\dot{\rho}$$ in Equation (5):
(12)$\dot{\epsilon_v^c} = {{1}\over{\rho}} B_0 \left[ 1 - e^{-B_1 \sigma} \right] e^{B_2 \rho}$
In the this implementation, it is assumed that volumetric compaction can only take place if the mean stress is compressive. Furthermore, a cap is assumed for the preceding expression so that no further compaction arises once the intact salt density has been reached.
Viscous Compaction
The total compaction strain rate has the expression
(13)$\dot\epsilon_{ij}^c = \dot\epsilon_v^c \left[ {{\delta_{ij}}\over{3}} - \beta {{{\sigma_{ik}^d}{\delta_{kj}}}\over{\bar{\sigma}}} \right]$
where $$\sigma_{ij}^d$$ is the deviatoric stress tensor, $$\bar\sigma = \sqrt{3J_2}$$ is the von Mises stress, and $$J_2 = \sigma_{ij}^d\sigma_{ij}^d / 2$$. In this formula, the parameter $$\beta$$ is a constant set equal to one, so that in a uniaxial compression test, the lateral compaction strain-rate components vanish.
Viscous Shear
The viscous shear strain rate corresponds to that of the WIPP model. The primary creep strain rate is the same as that given in the WIPP model, but the secondary creep strain-rate has the deviatoric stress magnitude, $$\bar\sigma$$, divided by the fractional density (Equation (6)). It has the form
(14)$\dot\epsilon_s=D\,{\left( {{\bar\sigma}\over{F_d}} \right)}^n e^{(-Q/RT)}$
where the parameters are as previously defined.
As the material approaches full compaction, the fractional density approaches one. Because a cap is introduced to eliminate further creep compaction when the intact salt density is reached, the viscous shear behavior evolves toward that of the intact salt. Note that in the framework of the WIPP model, the intact salt creep behavior is triggered by deviatoric stresses, while the volumetric behavior is elastic.
Implementation
In the this implementation of the crushed-salt model, the total stresses and strain rates are decomposed into volumetric and deviatoric components. The incremental equations governing the volumetric behavior are linearized and solved explicitly for the mean stress increment. The creep compaction strain-rate is then derived and used in the expression for the deviatoric behavior whose implementation otherwise closely follows that adopted for the WIPP model. Finally, total stresses for the step are evaluated from the updated volumetric and deviatoric components.
References
Callahan, G.D., and K.L. DeVries. Analysis of Backfilled Transuranic Waste Storage Rooms, RE/SPEC, Inc., report to Sandia National Laboratories SAND91-7052 (1991).
Sjaardema, G.D., and R.D. Krieg. A Constitutive Model for the Consolidation of WIPP Crushed Salt and Its Use in Analyses of Backfilled Shaft and Drift Configurations, Sandia National Laboratories, SAND87-1977 (1987).
wipp-salt Model Properties
Use the following keywords with the zone property (FLAC3D) or zone property (3DEC) command to set these properties of the WIPP-salt model.
wipp-salt
activation-energy f
activation energy, $$Q$$
bulk f
bulk modulus, $$K$$
bulk-final f
final, intact salt, bulk modulus, $$K_f$$
constant-a f
WIPP model constant, $$A$$
constant-b f
WIPP model constant, $$B$$
constant-d f
WIPP model constant, $$D$$
constant-gas f
gas constant, $$R$$
compaction-0 f
creep compaction parameter, $$B_0$$
compaction-1 f
creep compaction parameter, $$B_1$$
compaction-2 f
creep compaction parameter, $$B_2$$
creep-rate-critical f
critical steady-state creep rate, $$ε̇^*_{ss}$$
density-final f
final, intact salt, density, $$ρ_f$$
density-salt f
density, $$ρ$$
exponent f
WIPP model exponent, $$n$$
poisson f
Poisson’s ratio, $$v$$
shear f
elastic shear modulus, $$G$$
shear-final f
final, intact salt, shear modulus, $$G_f$$
temperature f
zone temperature, $$T$$
young f
Young’s modulus, $$E$$
compaction-bulk f (r)
creep compaction parameter, $$K$$
compaction-shear f (r)
creep compaction parameter, $$G$$
density-fractional f (r)
current fractional density, $$F_d$$
Key
• Only one of the two options is required to define the elasticity: bulk modulus $$K$$ and shear modulus $$G$$, or Young’s modulus $$E$$ and Poisson’s ratio $$v$$.
• The two densities: density-salt $$ρ$$, and density-final $$ρ_f$$ , should be considered as internal variables for the constitutive level calculation only, which should not be confused with the real material density. | 2,275 | 8,228 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-49 | latest | en | 0.869354 |
http://www.slideserve.com/bernad/zuhone-flash | 1,493,306,353,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122174.32/warc/CC-MAIN-20170423031202-00631-ip-10-145-167-34.ec2.internal.warc.gz | 679,986,073 | 19,963 | # Novae and Mixing - PowerPoint PPT Presentation
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Novae and Mixing John ZuHone ASCI/Alliances Center for Thermonuclear Flashes University of Chicago Overview Purpose What is FLASH? Mixing in Novae Setting Up the Problem Doing the Problem Conclusions Purpose
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## Novae and Mixing
John ZuHone
ASCI/Alliances Center for Thermonuclear Flashes
University of Chicago
### Overview
• Purpose
• What is FLASH?
• Mixing in Novae
• Setting Up the Problem
• Doing the Problem
• Conclusions
### Purpose
• To develop a numerical simulation using the FLASH code to simulate mixing of flulds at the surface of a white dwarf star
• Understanding this mixing will contribute to our understanding of novae explosions in binary systems containing a white dwarf star
### What is FLASH?
• FLASH is a three dimensional hydrodynamics code that solves the Euler equations of hydrodynamics
• FLASH uses an adaptive mesh of points that can adjust to areas of the grid that need more refinement for increased accuracy
• FLASH also can account for other physics, such as nuclear reactions and gravity
### What is FLASH?
• Euler equations of hydrodynamics
¶r/¶t + Ñ • rv = 0
¶rv/¶t + Ñ • rvv + ÑP = rg
¶rE/¶t + Ñ • (rE + P) v = rv • g
where
E = e + ½v2
### What is FLASH?
• Pressure obtained using equation of state
• ideal gas
P = (g- 1)re
• other equations of state (i.e. for degenerate Fermi gases, radiation, etc.)
• For reactive flows track each species
¶rXl/¶t + Ñ • rXlv = 0
### Mixing in Novae
• What is a nova?
• novae occur in binary star systems consisting of a white dwarf star and a companion star
• the white dwarf accretes material into an accretion disk around it from the companion
• some of this material ends up in a H-He envelope on the surface of the white dwarf
### Mixing in Novae
• this material gets heated and compressed by the action of gravity
• at the base of this layer, turbulent mixing mixes the stellar composition with the white dwarf composition (C, N, O, etc.)
• temperatures and pressures are driven high enough for thermonuclear runaway to occur (via the CNO cycle) and the radiation causes the brightness increase and blows the layer off
### Setting Up the Problem
• Initial Conditions
• what we want is a stable model of a white dwarf star and an accretion envelope in hydrostatic equilibrium
• we get close enough to the surface where Cartesian coordinates (x, y, z) and a constant gravitational field are valid approximations
### Setting Up the Problem
• Hydrostatic Equilibrium
• to ensure a stable solution we must set up the initial model to be in hydrostatic equilibrium, meaning v = 0 everywhere
• momentum equation reduces to
ÑP = rg
• set this up using finite difference method, taking an average of densities
### Setting Up the Problem
• Procedure for initial model
• set a density at the interface
• set temperature, elemental abundances
• call equation of state to get pressure
• iterate hydrostatic equilibrium condition and equation of state to get pressure, density, etc. in rest of domain
### Setting Up the Problem
• Region I: 50% C, 50% O, T1 = 107 K
• Region II: 75% H, 25% He, T2 = 108 K
• Density at interface:
ro = 3.4 × 103 g cm-3
### Doing the Problem
• load the model in and see if the simulation is in hydrostatic equilibrium
• it ISN’T!
• high velocities at interface and boundary
• begin to examine the model for possible flaws
### Doing the Problem
• Question: Is the model itself really in hydrostatic equilibrium?
• test the condition, discover that the model is in hydrostatic equilibrium to about one part in 1012
• Question: Is the resolution high enough?
• try increasing number of points read in, increase refinement, still no change
### Doing the Problem
• Question: Is the density jump across the interface hurting accuracy?
• smooth out density jump by linearly changing temperature and abundances
• velocities slightly lower, but still present
• try this for a number of different sizes of smoothing regions, still no change
### Doing the Problem
• Check the equation of state
• the Helmholtz equation of state we were using was complex
• accounts for gas, degenerate electrons, and radiation
• switch to gamma equation of state to see if anything improves
• NO IMPROVEMENT!
## MovieTime!
(maybe)
### Doing the Problem
• Two important resolutions
• there was an error in temperature calculation which was caused by a mismatch in precision of numerical constants
• we found that if we used the same number of points in FLASH as we did the initial model some of the inconsistency was resolved
### Doing the Problem
• Which brings us to where we currently are…
• we believe that by our linear interpolation for the density is too imprecise
• we are currently implementing a quadratic interpolation for the density
### Conclusions
• What have we learned?
• stability is important
• the need for there to be a check within FLASH itself for hydrostatic equilibrium
• the need to carefully examine all parts of a code to look for possible mistakes
• consistency!
### Conclusions
• Thanks to:
• Mike Zingale and Jonathan Dursi
• Prof. Don Lamb
• the ASCI FLASH Center
• the University of Chicago | 1,377 | 5,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-17 | longest | en | 0.862321 |
http://ask.cvxr.com/t/concave-function/4813 | 1,544,526,569,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823618.14/warc/CC-MAIN-20181211104429-20181211125929-00603.warc.gz | 27,068,533 | 2,749 | # Concave function?
(Chethan Kumar Anjinappa) #1
Hi All,
Is the function `f(x,y,z) = y*log(1+(c*x/(y*z)))` is concave ? If so, how can I implement it in cvx (using rel_entr)?
with x,y,z>=0
(Mark L. Stone) #2
f(x,y,z) is neither convex nor concave. For instance, at x = y = z = 1, its Hessian has 2 negative eigenvalues and 1 positive eigemvalue, so is indefinite there.
If z were a constant, not a (CVX) variable, f could be expressed as `-rel_entr(y,y+c*x/z)` and would be concave as a function of x and y. | 169 | 514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-51 | latest | en | 0.897466 |
http://www.ck12.org/arithmetic/Problem-Solving-Plan-Diagrams/lesson/Problem-Solving-Plan-Diagrams/r11/ | 1,455,358,745,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701166261.11/warc/CC-MAIN-20160205193926-00253-ip-10-236-182-209.ec2.internal.warc.gz | 341,016,973 | 39,363 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Problem Solving Plan, Diagrams
## Develop selected strategies for problem solving.
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Problem Solving Plan, Diagrams
On his last day with Uncle Larry, Travis worked with Mr. Wilson on laying tile on the kitchen floor. Travis worked hard all morning and he was a bit discouraged when he reached his first break and realized that he had only finished about one-third of the floor. It had taken Travis two hours to tile one-third of the floor. He thought about this as he drank from his water bottle and ate an apple. “If it took me this long to tile one-third, how long will it take me to finish?” Travis wondered.
The floor is divided into 12 sections. If he has finished one-third of them, how many sections has he completed? This is the number that he completed in the two hours.
How many sections does he have left to complete? About how long will it take him to finish the rest?
There are many different strategies you could use to help Travis solve this problem, but drawing a diagram is probably the most useful. This Concept will show you how to effectively use a diagram to solve a problem.
### Guidance
You have been learning about fractions and mixed numbers and about how to add and subtract them. Many of the examples in these Concepts have used pictures to help you learn to solve them. Drawing a diagram or a picture is a strategy to help you solve many different problems. The first thing that you have to do when approaching a problem is to read and understand the problem and how to solve it.
John ate 15\begin{align*}\frac{1}{5}\end{align*} of the cake. What fraction is left?
First, you can see that we have the amount of cake the John ate and we need to know how much he has left. We are going to be subtracting. Let’s draw a diagram to show what we know about John and his cake.
Now that we have looked at what we know and what we need to know, we can draw the diagram. This is a diagram of fraction bars to represent John’s cake. The blue section shows how much of the cake John has eaten. The white bars represent the amount of cake that is left.
Here is the one-fifth that John ate. You can see that there are four-fifths left.
The answer to the problem is four-fifths.
Sometimes, we can set up a problem as addition and sometimes we can set it up as subtraction. Often times both ways will work but one will make more sense than the other.
Shannon jogged 1320\begin{align*}1 \frac{3}{20}\end{align*} miles yesterday. Today, she jogged 12\begin{align*}\frac{1}{2}\end{align*} mile. How many total miles did Shannon jog?
Method one –– Draw a diagram:
One way to solve this problem is to draw a diagram. Let’s start by looking at the first distance that Shannon jogged. Draw two same-sized rectangles. Divide one rectangle into 20 equal-sized sections. Then shade 1320\begin{align*}1 \frac{3}{20}\end{align*} of the diagram.
This represents the 1320\begin{align*}1 \frac{3}{20}\end{align*} miles that Shannon jogged yesterday.
Shannon also jogged 12\begin{align*}\frac{1}{2}\end{align*} mile today.
So, shade 12\begin{align*}\frac{1}{2}\end{align*} of the partially filled rectangle to represent the distance she jogged today.
The diagram is 11320\begin{align*}1 \frac{13}{20}\end{align*} shaded. So, Shannon jogged a total of 11320\begin{align*}1 \frac{13}{20}\end{align*} miles on those two days.
Method two –– Set up an addition problem:
To find out how many miles she jogged all together, add 1320+12\begin{align*}1 \frac{3}{20} + \frac{1}{2}\end{align*}. The fractional part of the mixed number has a different denominator than 12\begin{align*}\frac{1}{2}\end{align*}. Find the least common multiple (LCM) of both denominators. The least common multiple of 20 and 2 is 20. Next, we rename the fractions.
12=1020
Now we can add the two together.
1320+1020=11320
Notice that our answer is the same. Both methods will produce the same result. You can choose the method that you find easiest when working on problems like this.
Now it's time for you to try a few. Draw a diagram and solve each problem.
#### Example A
213+4\begin{align*}2 \frac{1}{3} + 4\end{align*}
Solution:613\begin{align*}6 \frac{1}{3}\end{align*}
#### Example B
4515\begin{align*}\frac{4}{5} - \frac{1}{5}\end{align*}
Solution:35\begin{align*}\frac{3}{5}\end{align*}
#### Example C
34+24\begin{align*}\frac{3}{4} + \frac{2}{4}\end{align*}
Solution:114\begin{align*}1 \frac{1}{4}\end{align*}
Let’s use a diagram to help Travis with his tiling project. Here is the original problem once again.
On his last day with Uncle Larry, Travis worked with Mr. Wilson on laying tile on the kitchen floor. Travis worked hard all morning and he was a bit discouraged when he reached his first break and realized that he had only finished about one-third of the floor. It had taken Travis two hours to tile one-third of the floor. He thought about this as he drank from his water bottle and ate an apple. “If it took me this long to tile one-third, how long will it take me to finish?” Travis wondered. The floor is divided into 12 sections. If he has finished one-third of them, how many sections has he completed? This is the number that he completed in the two hours. How many sections does he have left to complete? About how long will it take him to finish the rest?
First, let’s underline all of the important information to help us read and understand the problem. Let’s figure out how much of the floor Travis has finished. First, let’s find an equivalent fraction for one-third with a denominator of 12.
13=412
Next, we can draw a diagram of the finished part of the floor.
Here is a picture of what Travis has finished.
How much does he have left?
We can count the units and see that he has 812\begin{align*}\frac{8}{12}\end{align*} of the floor left to tile. This is double what he did in two hours.
Travis has about four hours of work left.
Travis finishes his break and gets back to work. If he continues working at the same pace, he will finish working around 2 pm just in time for some pizza for lunch.
### Vocabulary
Problem Solving
using key words and operations to solve mathematical dilemmas written in verbal language
Diagram
a drawing used to represent a mathematical problem.
### Guided Practice
Here is one for you to try on your own.
Teri ran 112\begin{align*}1 \frac{1}{2}\end{align*} miles yesterday, and she ran 212\begin{align*}2 \frac{1}{2}\end{align*} miles today. How many miles did she run in all?
If John ran 7 miles, what is the difference between his total miles and Teri’s total miles? How many miles have they run altogether?
We can solve this problem a couple of different ways. First, we could draw a diagram of the path of both runners.
112+212=4\begin{align*}1 \frac{1}{2} + 2 \frac{1}{2} = 4\end{align*} miles
Terri ran 4 miles.
John ran 7 miles.
There is a difference of 3 miles.
Together, they ran 11 miles.
### Practice
Directions: Solve each of the following problems by using a problem solving strategy.
1. Tyler has eaten one-fifth of the pizza. If he eats another two-fifths of the pizza, what part of the pizza does he have left?
2. What part has he eaten in all?
3. How many parts of this pizza make a whole?
4. Maria decides to join Tyler in eating pizza. She orders a vegetarian pizza with six slices. If she eats two slices of pizza, what fraction has she eaten?
5. What fraction does she have left?
6. If Tyler was to eat half of Maria’s pizza, how many pieces would that be?
7. If Maria eats one-third, and Tyler eats half, what fraction of the pizza is left?
8. How much of the pizza have they eaten altogether?
9. John and Terri each ran 18 miles. If Kyle ran half the distance that both John and Teri ran, how many miles did he run?
10. If Jeff ran 312\begin{align*}3 \frac{1}{2}\end{align*} miles, how much did he and Kyle run altogether?
11. What is the distance between Jeff and Kyle’s combined mileage and John and Teri’s combined mileage?
12. Sarah gave Joey one-third of the pie. Kara gave him one-fourth of another pie. How much pie did Joey receive altogether?
13. Is this less than or more than one-half of a pie?
14. Who gave Joey a larger part of the pie, Kara or Sarah?
15. What is the difference between the two fractions of pie?
### Vocabulary Language: English
Diagram
Diagram
A diagram is a drawing used to represent a mathematical problem.
Problem Solving
Problem Solving
Problem solving is using key words and operations to solve mathematical dilemmas written in verbal language. | 2,253 | 8,843 | {"found_math": true, "script_math_tex": 22, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2016-07 | latest | en | 0.966313 |
https://codeproject.freetls.fastly.net/Articles/706136/Csharp-Bin-Packing-Cutting-Stock-Solver?msg=5722453 | 1,631,969,236,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00304.warc.gz | 226,116,508 | 32,300 | 15,030,744 members
Articles / General Programming / Algorithms
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Posted 5 Jan 2014
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# C# Bin Packing - Cutting Stock Solver
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Application for solving Bin Packing and Cutting Stock problem
## Introduction
The original idea behind this article is a proposal for the solution of two well known classes of Operation Research problems: the Bin Packing and the Cutting Stock with many sizes of Bin/Stock. Altough this work meets the traditional one dimension problem, the exposed approach could be employed to solve the two and three dimensions problem. This work also include some useful features, such as the possibility of considering cost and quantities constraints in searching solutions. Finally, remaining in the very spirit of CodeProject, two more non- design time features has been added: constraints on minimum reject, and a method for improving solutions by making "local moves" - see the History section below and the related Comments and Discussion section.
## Background
In the Bin Packing Problem a set of items must be grouped into bins of a certain size (capacity of the bin): the aim is to minimize the total number of bins.
In the Cutting Stock Problem a set of items must be cutted from Stocks of a certain size (length of the stock): the aim is to minimize the total number of stocks.
In the past several different methods were developed by researcher in order to solve this couple of very interesting and practical problems, including linear programming, heuristics, and evolutionary algorithms such as genetic algorithms and ant colony optimization systems [1, 2, 3, 4, 5]. Being known that in Operation Research only small size problems can be exhaustively investigated in order to find the "Absolute optimum" solution, it's obvious that any solver attempt aims to find a more general "Optimum solution" which only sometimes can be also an "Absolute optimum".
To complicate things it should be noticed that in practice it's fairly common to tackle problems with bins of multiple sizes, or stocks of multiple length. This leads to an exponential growth of the problem size.
It should be clear by definitions given that the Bin Packing Problem and the Cutting Stock Problem share a common nature. The algorithm proposed in this work is able to handle both of them.
## Using the application
To understand how this application works we follow an example that I took from
http://www.codeproject.com/Questions/553711/optimizedpluscuttingplussolutionplus-knapsack-2fbi
For simplicity, I write directly here the question:
Quote:
Hi,
I need the following solution and have been battling to get it right. I have also searched high and low but could not find anything that I could use. Perhaps I am not understanding my own problem well enough.
I need to be able to programmatically find the best way to cut lengths of Rope/Pipe.
Here is the scenario:
I have the following lenthgs in stock: 3 X 4.7m 5 X 7.4m 3 X 11.2m 4 X 9.8m (these are lengths of rope that I currently have in stock).
I need to cut the following lengths for an order: 2 X 3m 4 X 1.2m 1 X 6m 2 X 5.4m
The application exploits a Windows Form which is suitable for a comfortable data input. In this example, at the original problem has been added dummies constraints on minimum rejects:
The Form exposes two `DataGridView` with their respective `BindingNavigators` and some `Buttons`
• SEARCH, which starts the solving process for the current problem
• RESET, which clears both the DataGridView for the set up of a new problem
• OPEN, which load a file saved problem
• SAVE, which save the current input data into a file
There is also a `CheckBox `with a red text that will appear at some point of the researching process, and in the lowest part of the Form there is a `ToolStrip` which brings a `ProgressBar` and a `Label` that give information about what is happening.
In the top-right part of the Form a message in the downright spirit of CodeProject can be red.
The user should insert the size and the number of pieces in the `DataGridViewItems`.
By default, if there isn’t explicit input by the user, the program will automatically consider a minimum of 1 piece for the just inserted item. A `TextBox` embedded in the `BindingNavigatorItems` will show the real-time total sum of the items.
The `DataGridViewStocks` should be filled with the size, the cost and the maximum number of available pieces for each stock (if it’s limited). To effectively consider this upper limit in the calculation a confirm by `CheckBox` is required. This is why it’s possible to simulate different scenarios simply checking or unchecking checkboxes.
Be aware that only the stock size is always taken into consideration during the algorithm, as will be clear after seen the flow chart.
One more thing is about the cost. Obviously this is the cost of the stocks, but for further analysis this value could include other parameters, such as cost for shipment and so on. In this way, a same size of stocks can be evaluated with two different costs and maximum number of pieces.
In the driver example written above, and downloadable at the top of the article, there isn’t information regarding the costs, so I fixed them in order to show how many result can be found.
However when the input is accomplished the “SEARCH” button must be pressed, and the program runs according to the following flow chart:
The First step is to build a quick solution, regardless the cost or the limited number of available pieces.
This solution has name “Bound Solution” and will be an upper bound for any further attempt of improvement.
The second step consists in a deeper look for a better solution, and it’s now that come into play the ‘optional parameters’ cost and max number of pieces.
As one can see running the application with the given input data, at this point our driver example finds two solution: one is a “Best Cost Solution”, while the other is a “Best Size Solution”.
Now if the problem is enough small and if the “Absolute Best Size Solution” (mathematically speaking regarding the total size of the solutions found) hasn’t been met, a further search is performed. The aim of this final step is to fit one of the many bin/stock combinations which have minimum sum. There is no guarantee that such a solution exists, and this process can be time consuming. At this point appears the above mentioned `CheckBox` with red text “End Search”, as the picture below shows. When checked, the search algorithm stops and results are immediately shown.
This snapshot gives an idea of the result:
:
By the way, for the driver example doesn’t exist any “Absolute Best Size Solution”. Howewer, if you are curious to see this event happen, you can slightly modify the item of size 5,4: change its size into 5,8 and ask for only 1 piece of it, then press “SEARCH” and wait a few seconds.
## A look to the algorithms
When the “SEARCH” `Button` is pressed, a CuttingStock object is constructed and a solving process begins according to the previous flow chart .
The Bound Solution is obtained exploiting a “Greedy First Fit” algorithm. In simple word, this algorithm processes all the items and assigns them to the biggest bin/stock available. When the bin/stock can’t hold the item, it tries to assign this item at the first bin/stock that can hold it. If no bin/stock has enough space, than a new bin/stock is used. After the assignement procedure the algorithm look for an improvement following these ordered steps:
1. Trying to reduce the size of some bin/stock if it's employ is less or equal the size of another bin/stock available. This is performed calling the DownSize(List<Bin> mySolution) method.
2. Trying to qualify the solution: the QualifySolution(List<Bin> mySolution) method is called, and it explores the possibility to free some space in the bin/stock making a series of local moves. The following, short paragraph will give more details about this idea.
3. Trying again to DownSize the solution .
Once got a Bound Solution, the application build a set of potential improving solution exploiting the BranchAndBound class. This class is inspired to the Branch & Bound strategy, which is fairly common in Operation Research and gives back a `List` of potential better solutions which is both sorted on Cost and on Size and is passed to a “Greedy Best Fit” algorithm until new solutions will be found. This greedy algorithm works in a similar way to the “Greedy First Fit”, the main differences being that now it tries to fit the items with the set derived by the Branch & Bound and that when a bin/stock can’t hold the item, it tries to assign this item at the bin/stock which can hold it giving the lowest reject. Hower, if it runs succesfully, it is able to return a so called "Best Size Solution" and "Best Cost Solution", both of them being precessed by the already told QualifySolution(List<Bin> mySolution) method.
At this point a new search will be performed only to achieve the Absolute Best Size with a "Greedy Next Fit" algorithm applied to the branches of the Branch & Bound space with size equal to the "Absolute optimum" value. This step will not be executed if "Best Size Solution" is also "Absolute Best", or if there are too much items to process (the design choice is 12). This greedy algorithm is the simplest and the quickest of all: the items are placed in a bin/stock until it’s enough empty, than a new bin/stock is added. However, to be sure every possible permutation of the items in ListOfItems must be tried to fit a given set of bins/stocks. This is the main reason for which I made the choice to introduce a constraint on the number of items to process. This seeming simple idea took several lines of code, and I leave the comments in it to better clarify the concept.
## What is a Qualified Solution?
According to the statements given in the Background section, the aim of the Bin Packing or Cutting Stock problem is to minimize the amount of row material - bins or stocks. Provided that we are able to get this purpose, we can consider a more advanced target: to get a solution which allow us a good reusability of its reject. Given a problem and its solutions, provided that they have the same total size of bins/stocks, we can consider as "more qualified" a solution which gives a reject still reusable in a next job process. At the opposite, we can consider as "less qualified" a solution whose reject is not reusable and so it's completely useless. Altough this aspect seems to be interesting, I haven’t found references to it in the work coming from the field of research. At the same way, I didn’t find any attention in optimizing according to the “Cost” or giving a limitation at the number of some stock - the two features I decided to set in the application at design time. To fit the need to get a "more qualified" solution the dedicated method QualifySolution(List<Bin> mySolution) has been added to the code. The basic plan behind this action in "qualifying" is to free space in the bins/stocks trying to move items from one bin/stock to another: this one is chosen with a "Best Fit" criteria, whilst the items to move are selected from the greatest to the lowest running back from the last bin/stock to the first bin/stock in the examined solution.
## Is it a good idea to ask for a minimum rejects?
As I said in the introduction, the main purpose of this article is to find a good, and possibly quick, solution at the traditional Bin Packing/Cutting Stock Problem, whose solving time may be unacceptable even for small set of bin/stocks and small number of items. This concept implies the objective function is the minimization of the row materials employed. Thus we can consider as good choices all that ideas which doesn't increase the row material consumption - regardless to what could happen in the next job, or in the next of the next... Inspired to this principle is the "local move" that I called "QualifySolution". The constraint of "Minimum Reject" could lead to a not worst solution. In this case it would be another good "local moves" for our job. On the other hand, more often the result could be worst. In this case, I suggestat least to compare this results with those obtained excluding the minimum rejects. Here is a point where readers and users can make their experiences, possibly giving reasonable feedback at the CodeProject community through the Comments and Discussion section.
## Points of Interest
This is my first C# application, and I must say that although I’m not at all a software developer I met several interesting feature in C# and .NET framework which I hope could be useful for other beginners:
• The use of `DataGridView `control with in-memory data structure instead of a physical database;
• The implementation of several sorting method for the `List` collections;
• The useful `Linq` query and the method to put the query result in a separated collection;
• The powerful couple of classes `BinaryFormatter` and` FileStream `which make simple storing in one file many collections of objects.
## References
[1] P.C. Gilmore, R.E. Gomory. A linear programming approach to the cutting stock problem. Operations Research, 9:848-859, 1961.
[2] Silvano Martello, Paolo Toth. Knapsack Problems, Algorithms and Computer Implementations. John Wiley and Sons Ltd., England, 1990.
[3] Emanuel Falkenauer, Alain Delchambre. A genetic algorithm for bin packing and line balancing. Proceedings of the IEEE 1992 International Conference on Robotics and Automation, Nice, France, May 1992.
[4] Emanuel Falkenauer. A hybrid grouping genetic algorithm for bin packing. Journal of Heuristics, 2:5-30, 1996.
[5] Rhyd Lewis. A General-Purpose Hill-Climbing Method for Order Independent Minimum grouping Problems: A Case Study in Graph Colouring and Bin Packing. Computers and Operations Research, vol. 36(7), pp. 2295-2310.
## History
• 2014-01-04
• Original article
• 2014-01-06
• 2014-08-02
• Modified both the article and the code due to the post dated 25-May-2014:
• Updated article
## Share
Engineer Italy
I am an Engineer experienced in designing instruments and controls for industrial polluted air and V.O.C treatment plants. Currently I work in designing radiant heating/cooling systems and related controls, solar systems and other solutions for buildings energy efficiency.
My first personal computer was a ZX Spectrum Sinclair that in the early ‘80s I used so much in playing games and learning the bases of Basic. Then in my career I met other languages. Other than programming my hobbies are graphology, acoustic guitar, good books, my orchard, astronomy and play chess.
I live in the North West of Italy, very close to mountains and lakes and I like simple things. In summer you can find me somewhere walking on the mountains or going around with my sport bike. In winter you can probably meet me in my chessclub.
If you are planning to visit Italy I will be glad to show you some of this beautiful places and landscapes.
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source code Matin Pirasteh26-Jul-21 2:31 Matin Pirasteh 26-Jul-21 2:31
source code Member 1278414925-Jul-21 22:45 Member 12784149 25-Jul-21 22:45
source code AlexPentagon1-Jun-21 6:15 AlexPentagon 1-Jun-21 6:15
source code Member 31451707-Apr-21 23:58 Member 3145170 7-Apr-21 23:58
Source Code Member 1479207927-Dec-20 0:44 Member 14792079 27-Dec-20 0:44
I would like the code in c # andreire23-Nov-20 5:54 andreire 23-Nov-20 5:54
ciao Alberto mi manderesti il codice ? andreire22-Nov-20 22:10 andreire 22-Nov-20 22:10
Re: ciao Alberto mi manderesti il codice ? OriginalGriff22-Nov-20 22:13 OriginalGriff 22-Nov-20 22:13
Source Code please shanojscp20-Oct-20 4:34 shanojscp 20-Oct-20 4:34
delete it! db_developer14-May-20 21:01 db_developer 14-May-20 21:01
My vote of 1 db_developer14-May-20 20:58 db_developer 14-May-20 20:58
This resource supposes to post a SOURCE CODE! I m looking for solution not something else. Possibly it is even not his app at all.
Source code Member 1420334617-Nov-19 3:50 Member 14203346 17-Nov-19 3:50
Source code Member 1156117628-Oct-19 4:52 Member 11561176 28-Oct-19 4:52
Source code Member 1106619316-Oct-19 14:47 Member 11066193 16-Oct-19 14:47
My vote of 5 Member 1453822722-Jul-19 10:59 Member 14538227 22-Jul-19 10:59
source Member 1453822722-Jul-19 10:36 Member 14538227 22-Jul-19 10:36
Hi Alberto allviewsystem from Bayamon15-May-19 9:26 allviewsystem from Bayamon 15-May-19 9:26
Re: Hi Alberto Member 145177741-Jul-19 9:22 Member 14517774 1-Jul-19 9:22
Source code aselan3-May-19 21:52 aselan 3-May-19 21:52
Source Code? Jeff.Bk3-May-19 0:04 Jeff.Bk 3-May-19 0:04
Source code? tampicoscott5-Mar-19 20:17 tampicoscott 5-Mar-19 20:17
Re: Source code? Member 145177741-Jul-19 9:23 Member 14517774 1-Jul-19 9:23
where is source code? Member 1358481814-Feb-19 17:35 Member 13584818 14-Feb-19 17:35
Source Code Member 1231561128-Jan-19 5:34 Member 12315611 28-Jan-19 5:34
No sources available wwwings17-Jan-18 5:35 wwwings 17-Jan-18 5:35
Last Visit: 31-Dec-99 18:00 Last Update: 18-Sep-21 2:47 Refresh 123 Next ᐅ | 4,096 | 17,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-39 | latest | en | 0.942619 |
https://matcomgrader.com/problem/9576/chinese-curves/ | 1,582,724,382,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146342.41/warc/CC-MAIN-20200226115522-20200226145522-00455.warc.gz | 448,520,309 | 52,917 | C - Chinese curves
Languages: C, C++, Java, Python, ... (details)
You are in a world with curves of the form $f(x) = \arctan(e ^ x + a) \sqrt{b \cdot x^2 + c}$, where $a$, $b$ and $c$ are integers. In this world, there are two types of queries:
• Given three integers $a$, $b$ and $c$, create a new curve with parameters $a$, $b$ and $c$.
• Given an integer $p$, determine the minimum value of $f(p)$ among all curves.
The following image corresponds to the sample test case, where the blue curve is the last curve added.
Here is an explanation of the sample input:
1. From the input, mask $m$ is intially set to $0$.
2. Query 1 creates a curve with parameter $a = 1000$, $b = 1$, $c = 1$ (since mask $m = 0$).
3. Query 2 asks for the minimum value of $f(1)$ among all curves. Since there is only one curve, answer is simply $\arctan(e ^ 1 + 1000) \sqrt{1^2 + 1}$, which is about $2.22$.
4. After answering query 2, mask $m$ becomes $\lfloor 2.22 \rfloor \oplus 0 = 2$.
5. Query 3 creates a curve with parameters $a$, $b$, and $c$ equal to $3 \oplus 2 = 1$.
6. As $p = 3 \oplus 2 = 1$, query 4 asks for the minimum value of $f(1)$ among the two existing curves. From the image, we can see that the last added curve yields a minimum value of about $1.85$.
7. After answering query 4, mask $m$ is updated to $\lfloor 1.85 \rfloor \oplus 2 = 3$.
8. The last query asks to find the minimum value of $f(41 \oplus 3) = f(42)$ among the two curves.
Input
In the first line, there are two integers $q$ and $m$, where $q$ is the number of queries to follow $(1 \le q \le 10 ^ 5)$ and $m$ is a mask $(0 \le m \le 10 ^ 5)$. Each of the following $q$ lines contains a query using one of these two formats:
• $1$ $a$ $b$ $c$: if the query is of the first type, where $0 \le b \le 10 ^ 6$, $0 \le a, c \le 10 ^{18}$.
• $2$ $p$: if the query is of the second type, where $0 \le p \le 10 ^ 6$.
In the input, the values of $a$, $b$, $c$ and $p$ are not given directly. Instead, you should restore them from the inputs $a'$, $b'$, $c'$ and $p'$ by computing $a = a' \oplus m$, $b = b' \oplus m$, $c = c' \oplus m$, and $p = p' \oplus m$, where $m$ is the current value of mask $m$ and $\oplus$ is the bitwise xor operator (available as the ^ symbol in many programming languages).
After each query of type 2, the mask $m$ changes to $\lfloor s \rfloor \oplus m$, where $s$ is the answer to the query, $\oplus$ is the bitwise xor operator, and $\lfloor s \rfloor$ is the floor function that yields the integer part of number $s$.
It is guaranteed that the first query is of type 1.
Output
For each query of type 2, output a line with the required answer, rounding to two digits of precision after the decimal point.
Sample test(s)
Input
5 0 1 1000 1 1 2 1 1 3 3 3 2 3 2 41
Output
2.22 1.85 65.99 | 937 | 2,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-10 | longest | en | 0.783241 |
https://aic-components.com/useful-articles/can-460-volts-be-single-phase.html | 1,638,318,100,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00047.warc.gz | 162,770,555 | 18,079 | # Can 460 volts be single phase?
Contents
## Is 460 volts the same as 480?
RE: Difference between 460V and 480V regarding…
480 volts is the system voltage. 460 volts is the rated voltage of your equipment. 460 volt rated equipment is always used on a 480 volt system to account for voltage drop due to starting or line losses.
## Is 440 volt the same as 460 volt?
460V was frequently used as the nameplate voltage for motors used on a 480V system. So a motor rated at 460V (3-phase) will work at 440V, indefinitely, without overheating or not delivering rated output.
## How many volts is a single-phase motor?
Common 60hz voltages for single-phase motors are 115 volt, 230 volt, and 115/230 volt. Common 60hz voltages for three-phase motors are 230 volt, 460 volt and 230/460 volt. Two hundred volt and 575 volt motors are sometimes encountered. In prior NEMA standards these voltages were listed as 208 or 220/440 or 550 volts.
## Is 480V always 3 phase?
480V can be classified as single and 3 phase circuits. 480V 3 phase circuits are the most common power systems used in US industrial plants and are considered to be low voltage power systems.
## Is 480 single phase?
Single phase 480 is 277 volt. To get 480 you need two phase at any given time to give you 480 volts.
THIS IS IMPORTANT: Which of the following code is used to give cutter offset compensation right of a part surface?
## Is 480 volts considered High Voltage?
Generac states that generators less than and equal to 600 volts are medium-voltage and generators greater than 600 volts are considered high voltage.
## Can a 460V motor run on 440V?
The current will adjust accordingly and as voltage drops the motor will become less efficient, but the motor will still deliver rated output. So a motor rated at 460V (3-phase) will work at 440V, indefinitely, without overheating or not delivering rated output.
## Can a 440V motor run on 480V?
A standard NEMA motor should be good for continuous operation at 110% of nameplate. Even if the voltage at the MCC is 480V, the voltage at the motor will be less. If these are old 440 V motors, they will probably be OK.
## Why 3 phase is called 440 volts not 660 volts?
In 3 phase supply, there are 3 supply lines phase shifted at 120 degrees from each other. So the net voltage difference between the two phases in accordance with the phase angle of 120 degrees is 440V. | 582 | 2,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-49 | latest | en | 0.886062 |
http://www.pharmtech.com/pharmtech/Analytics/Multifactor-Non-linear-Modeling-for-Accelerated-St/ArticleStandard/Article/detail/848456?contextCategoryId=50207 | 1,416,777,089,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400380037.17/warc/CC-MAIN-20141119123300-00033-ip-10-235-23-156.ec2.internal.warc.gz | 809,292,106 | 40,836 | Multifactor Non-linear Modeling for Accelerated Stability Analysis and Prediction - Pharmaceutical Technology
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Multifactor Non-linear Modeling for Accelerated Stability Analysis and Prediction
The right approach can provide a clear, statistically defendable method for determining dissolution and accelerated stability.
Jul 2, 2014 Pharmaceutical Technology Volume 38, Issue 7, pp. 46-49
Accelerated stability analysis is a strategy used to quickly evaluate alternative formulations, packaging, and processes. Accelerated linear studies are commonly performed and modeled; however, accelerated multiple-factor non-linear modeling has been a gap, and statistical software tools such as SAS/JMP do not directly have any provision to model multiple factor nonlinear responses. This paper outlines an approach to model and predict non-linear multiple factor stability/tablet dissolution data under accelerated and nominal storage conditions.
RELATED CONTENT Data Integrity in the Analytical Laboratory More Rapid Extractable & Leachable Analyses with Advanced Mass Spectrometers More Analytics
There are many non-linear stability cases such as dissolution, leachables, and moisture. Being able to model these non-linear processes is crucial to proper drug development. In addition to general non-linear modeling, there are multiple factors that may influence the non-linear curve. The following is a list of factors that may impact the asymptotes, growth rate, and inflection point of a curve:
• Stability storage temperature and humidity
• Particle size
• pH
• Amount of an excipient
• Processing conditions and/or set points
• Packaging materials/method
Study design
Proper design of experiments for data collection and curve isolation is crucial for building non-linear models. Figure 1 illustrates the factors that should all be square relative to all other factors and have zero correlation relative to each other. For this tablet dissolution example, multiple time points (minutes), multiple storage conditions (temperature), multiple drug substance particle sizes, and multiple weeks were measured. Percent dissolution was the response of interest.
Analysis method
The following is a step-by-step procedure for non-linear stability modeling and expiry determination.
Step one. Measure the data at multiple time periods, using multiple particle sizes and at multiple temperatures. Generate a plot of the data to visualize the relationship of the curves over time (25-10-0: 25=Temperature, 10=Particle Size, and 0 = days) (see
Figure 2).
Step two. Fit each curve individually. In this example, each dissolution curve was fit using a four-parameter logistics (4PL) curve. The four parameters are: upper asymptote, lower asymptote, inflection point, and slope of the dissolution curve. R-Square should be high (typically above 0.95) and RMSE error should be low for each curve. Outliers should be checked using the residuals. Save the parameters of the curve. In this example, there are four parameters, upper asymptote, lower asymptote, growth rate (slope), and inflection point (see Figure 3).
Step three. Save the parameters from the curve and the factors that influence them into a table. For this example, the growth rate and inflection point are the coefficients that may change the most based on the factors under consideration. The upper and lower asymptotes are not of concern for this problem as all of the curves have similar lower asymptotes and similar upper asymptotes, but upper and lower asymptotes could be important for other problems, so generally it is best to model all of the curve parameters (see Figure 4).
Step four. Fit the curve parameters with a least-squares multivariate regression. Growth rate, the slope of the 4PL fit, and inflection point are the most important as the dissolution starting point and the upper asymptote are essentially the same for all curves. Main effects and interaction models generally work best and p-values and F tests can be used to evaluate each model term (see Figure 5).
Step five. Save the equation from the multivariate parameter model. An example of the inflection point model is in Equation 1.
[Eq. 1]
Step six. Substitute the growth rate (slope) and inflection point coefficients from the multivariate model into the nonlinear prediction. The red box in Figure 6 shows the substitution for the inflection point.
Step seven. Check the model to make sure it matches the actual data. Correct any modeling errors by adding or modifying a multivariate model term. A simple YX regression plot of the model versus the measurement will indicate model quality and any systematic errors (see Figure 7).
Step eight. Create a profiler from the equation to predict future dissolution rates. This can be done using a modern statistical package such as SAS/JMP (see Figure 8).
Step nine. Predict dissolution at any time, temperature, or particle size combination using the profiler. For this example, particle size was set to 5 μm, temp to 25 and time in weeks to zero. The dissolution time was fixed at 100 min with a specification of not less than 90% (see Figure 9).
Step ten. From the profiler at each time point (weeks), make sure the time (min) and particle size (5 um) are fixed, determine dissolution at the nominal temperature (non-accelerated condition). Fit the rate of degradation using either a linear or nonlinear model from the profiler predicted data. In this case, the rate of dissolution is not linear so a non-linear curve is fit to the data and the expiry is determined based on the extrapolated curve (see Figure 10).
The same method is used for predicting both the nominal expiry and the 95% CI expiry.
Step eleven. Finally, long-term stability evaluation at nominal storage conditions will be used to confirm the early model prediction and will provide an independent secondary determination of stability and changes in dissolution. Understanding rates of change should factor into shelf life and release specification limits (1).
Summary
Non-linear multiple factor analysis has long been a problem in a variety of process and product modeling and prediction. The novel procedure discussed in this paper for the characterization of multiple factor non-linear product performance provides a clear, statistically defendable method for determining dissolution and accelerated stability. Long-term verification of accelerated conditions should always follow early determinations of expiry, acceleration rates, and rates of degradation.
Reference
1. ICH, Q6A Specifications: Test Procedures and Acceptance Criteria for New Drug Substances and New Drug Products: Chemical Substances (Oct. 6, 1999).
Mark Alasandro, PhD, is director, Allergan Irvine. Thomas A. Little PhD, is president, Thomas A. Little Consulting, drlittle@dr-tom.com.
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https://math.stackexchange.com/questions/727662/a-planar-brownian-motion-has-area-zero/732272 | 1,558,557,894,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256958.53/warc/CC-MAIN-20190522203319-20190522225319-00434.warc.gz | 556,828,021 | 42,541 | # A planar Brownian motion has area zero
I'm looking for proofs of Paul Lévy's theorem that a planar Brownian motion has Lebesgue measure $0$. I know of only two proofs: one is in Lévy's original paper (Théorème 12, p. 532) and the other is in Mörters & Peres's "Brownian Motion" (Theorem 2.24, p. 46).
Unfortunately, Mörters & Peres's proof leaves out many technical details, without which I find the proof hopelessly difficult to understand.
I have not attempted to read Lévy's original proof, partly because it's in French (though I can read French, with effort) and partly because it's in the middle of a long article that was not written for students, but rather for professional mathematicians, and hence I suspect that it would be a pain for me to read it.
I have not been able to find any other sources containing a proof of this theorem. If someone knows of any such sources, in English, French or German, ideally textbooks (but any other source will be helpful too), please let me know.
Thank you.
Unfortunately, I'm not aware of an alternative proof (or a more detailed version of the proof by Mörters/Peres). So, instead of providing you with a source, I'll follow the lines of Paul Lévy; hopefully filling the gaps in his proof.
Let $(B(t))_{t \geq 0}=((B^1(t),B^2(t)))_{t \geq 0}$ be a two-dimensional Brownian motion and denote by $\lambda$ the (two-dimensional) Lebesgue measure.
First, we show that $L := \lambda(B([0,1]))$ satisfies $\mathbb{E}L<\infty$. To this end, note that
\begin{align*} \{L>4r^2\} &\subseteq \left\{ \max_{t \in [0,1]}(|B^1(t)|,|B^2(t)|) > r \right\} \\ &= \left\{ \max_{t \in [0,1]} B^1(t) > r \right\} \cup \left\{ \max_{t \in [0,1]} B^2(t) > r \right\} \\ &\quad \cup \left\{ \min_{t \in [0,1]} B^1(t) < -r \right\}\cup \left\{ \min_{t \in [0,1]} B^2(t) < -r \right\}.\end{align*}
From the reflection principle we know that $\max_{t \in [0,1]} B^j(t) \sim -\min_{t \in [0,1]} B^j(t) \sim |B^j(1)|$ for $j=1,2$. Hence,
$$\mathbb{P}(L>4r^2) \leq 4 \sqrt{\frac{2}{\pi}} \int_r^{\infty} \exp \left( - \frac{y^2}{2} \right) \, dy \leq \frac{4 \sqrt{2}}{\pi} \frac{1}{r} \exp \left(- \frac{r^2}{2} \right).$$
Combining this estimate with the fact that $\mathbb{E}L = \int_0^{\infty} \mathbb{P}(L >r) \, dr$, we get $\mathbb{E}L<\infty$.
Next, we note that for the restarted Brownian motion $W_t := B_{t+1}-B_1$, we have $\lambda(W([0,1])) = \lambda(B([1,2]))$ and, since $W \sim B$, we conclude that $$\mathbb{E}(\lambda(B([0,1]))) = \mathbb{E}(\lambda(B([1,2]))).$$ Similarly, as $\tilde{W}_t := \frac{1}{\sqrt{2}} B_{2t}$ is a Brownian motion, we have
$$\lambda(\tilde{W}([0,1])) = \lambda \left( \frac{1}{\sqrt{2}} B([0,2]) \right) = \frac{1}{2} \lambda(B([0,2])$$
i.e. $\lambda(B([0,2])) \sim 2 \lambda(B([0,1]))=2L$. Therefore,
\begin{align*} 2\mathbb{E}L = \mathbb{E}(\lambda(B([0,2]) &= \mathbb{E}(\lambda(B([0,1])) + \mathbb{E}(\lambda(B([1,2])) - \mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))) \\ &= 2\mathbb{E}L + \mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))). \end{align*}
Hence, $$\mathbb{E}\big(\lambda(B([0,1]) \cap B([1,2]))\big)=0. \tag{1}$$
By Fubini's theorem, we have
$$\mathbb{E}L = \int \int 1_{B([0,1])}(x,y) d\lambda(x,y) \, d\mathbb{P} = \int \underbrace{\mathbb{P}((x,y) \in B([0,1]))}_{=:p(x,y)} \, d\lambda(x,y). \tag{2}$$
If we set $W_t := B_{1-t}-B_1$ and $\tilde{W}_t := B_{t+1}-B_1$, then both processes are Brownian motions and
$$W([0,1]) = B([0,1])-B_1 \qquad \qquad \tilde{W}([0,1]) = B([1,2])-B_1.$$
Using that $(W_t)_{t \geq 0}$ and $(\tilde{W}_t)_{t \geq 0}$ are independent, we get
\begin{align*}\mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))) &= \mathbb{E}(\lambda(W([0,1]) \cap \tilde{W}([0,1]))) \\ &= \int \mathbb{P}((x,y) \in W([0,1]) \cap \tilde{W}([0,1])) \, d\lambda(x,y) \\ &= \int \mathbb{P}((x,y) \in W([0,1])) \mathbb{P}((x,y) \in \tilde{W}([0,1])) \, d\lambda(x,y) \\ &= \int p^2(x,y) \, d\lambda(x,y). \end{align*}
Now $(1)$ implies $p(x,y)=0$ $\lambda$-almost everywhere and therefore, by $(2)$, $\mathbb{E}L=0$. Hence, as $L \geq 0$, we finally conclude $L=0$ almost surely.
Remark It is not obvious that $\omega \mapsto L(\omega)$ and $((x,y),\omega) \mapsto 1_{B([0,1],\omega)}((x,y))$ are random variables; see this question and Evan Aad's answer for the measurability of $L$ and this question for the measurability of the second mapping. (Note that measurability of $((x,y),\omega) \mapsto 1_{B([0,1],\omega)}((x,y)$ entails the measurability of $L$, by Fubini's theorem.)
• Thank you very much. Unfortunately, I won't be able to read your answer until tomorrow. – Evan Aad Mar 30 '14 at 12:19
• @EvanAad No problem, I'm not in a hurry. – saz Mar 30 '14 at 14:57
• Before I read your answer, I'd like to understand why $L$ is a random variable. I have read Lévy's proof on p. 533, but there are a couple points I didn't get. I have written then as another question here. – Evan Aad Apr 1 '14 at 6:55
• Do you understand why $L$ is a random variable? Cause I sure don't. – Evan Aad Apr 3 '14 at 19:05
• @EvanAad Sorry, I don't have that much time to think about it right now; I'll take a look at it this weekend. – saz Apr 3 '14 at 19:16
I will complement saz's answerby proving that $L$ (to use saz's terminology) is a random variable.
EDIT: A simpler and more powerful proof was given later by saz here (see his remark at the end).
Introduction & proof outline 0
Following is a proof that the measure of the path of an $n$-dimensional ($n \geq 1$) Brownian motion is itself a measurable function. The proof is based on ideas presented by Paul Lévy in [L] (p. 533), but mine is not exactly the same as the proof given there, simply because I didn't quite understand that proof.
I first establish some notations (Notation 1), which are used in the sequel. I then introduce three constants (Constants 2) that will be in effect throughout. A few definitions and lemmas lead to the main result (theorem 9). The functions introduced in Definition 5 and Definition 7 correspond, more or less, to objects used in Lévy's proof. Lemma 6 and Lemma 8 study the properties and inter-relationships between these objects.
The idea of the proof is to cover the Brownian path segment between the times $0$ and $t$ ($g_t(\omega)$) by two sets: $f_{m, r, t}(\omega)$ and $h_{r, t}(\omega)$. $f_{m, r, t}(\omega)$'s measure $F_{m, r, t}(\omega)$ can readily be shown to be a measurable function of $\omega$, whereas $h_{r, t}(\omega)$'s measure $H_{r, t}(\omega)$ can readily be shown to converge to $g_t(\omega)$'s measure $G_t(\omega)$, as $r$ diminishes to $0$. These two desireable properties ($F_{m, r, t}$'s measurability and $H_{r, t}$'s convergence to $G_t$) are combined (Theorem 9.1) using a result that I proved in another post (theorem 1 here).
The reason why $F_{m, r, t}$'s measurability is relatively easy to demonstrate, is that $f_{m, r, t}(\omega)$ has a simple structure: it is the union of a finite number ($m$) of balls of the same radius ($r$). Such unions are denoted below by $e_r(x_1, \dots, x_m)$ with $x_1, \dots, x_m$ marking the centers of the balls. To assist in the study of the properties of $e_r(x_1, \dots, x_m)$'s measure $E_r(x_1, \dots, x_m)$, we use bounds derived in Lemma 4.
All the functions listed in the previous two paragraphs are introduced in Definition 5 and studied in Lemma 6. The functions introduced in Definition 7, whose properties are studied in Lemma 8, are used in the main result (Theorem 9.1) to construct a sequence of covers $f_{m_k, 1 / k, t}$ converging to $g_t$ and to quantify the rate of convergence of $F_{m_k, 1/k, t}$ to $G_t$. To quantify this rate of convergence, a uniform upper bound ($u$) is calculated (Lemmas 8.2(c)) on the maximal distance ($d_{a, b}(\omega)$) between two points that lie on the path segment $g_t(\omega)$ between times $a$ and $b$.
The entire proof is carried out under the assumption (Constants 2.1) that $n \geq 2$. In the end (Remark 10) I indicate how to extend the proof to the case $n = 1$.
The published works referenced in the post are listed at the end of the post in the section "Works cited".
Notation 1
For every $n \in \mathbb{N}_1 := \{1, 2, \dots\}$,
1. Denote the Borel field on $\mathbb{R}^n$ by $\mathcal{B}_n$,
2. Denote the Lebesgue measure on $\mathcal{B}_n$ by $\lambda_n$ and the outer measure on $\mathbb{R}^n$ by $\lambda_n^*$,
3. Denote by $|\cdot|_n$ the standard Euclidean norm on $\mathbb{R}^n$,
4. For every $x \in \mathbb{R}^n$ and every $r \in (0,\infty)$, define \begin{align} \mathbf{B}_x(r) & := \{y \in \mathbb{R}^n \mid: |y - x|_n < r\} \\ \overline{\mathbf{B}_x(r)} & := \{y \in \mathbb{R}^n \mid: |y - x|_n \leq r\} \end{align}
5. Denote the topology induced by $|\cdot|_n$ on $\mathbb{R}^n$ by $\mathcal{O}_n$ and for every $m \in \mathbb{N}_1$, denote the product topology induced by $|\cdot|_n$ on $(\mathbb{R}^n)^m$ by $\mathcal{O}_{n;m}$.
6. If $S$ is a set, we denote its power set by $2^S$.
7. Denote by $\mathbf{C}_{(0; n)}$ the set of continuous functions $f : [0, \infty) \rightarrow \mathbb{R}^n$, such that $f(0) = 0$. Denote by $\mathcal{B}(\mathbf{C}_{(0; n)})$ the Borel $\sigma$-algebra on $\mathbf{C}_{(0; n)}$ (cf. [S], p. 41).
Constants 2
Fix the following for the remaining of this post.
1. $n \in \mathbb{N}_1$, such that $n \geq 2$,
2. $S := (\Omega, \mathcal{F}, P)$ - A probability space,
3. $W : \Omega \times [0,\infty) \rightarrow \mathbb{R}^n$ - a standard, $n$-dimensional Brownian motion over $S$. Denote $W$'s components thus: $W = (W_1, \dots, W_n)$.
Definition 3
Set $$c := \frac{\pi^{n / 2}}{\Gamma(\frac{n}{2} + 1)}$$ (see the next lemma for the rationale.)
Lemma 4
1. Let $x, y \in \mathbb{R}^n$, let $\varepsilon \in (0,1]$ and let $r \in (0,\infty)$. If $|x - y|_n < \varepsilon$, $$\lambda_n(\mathbf{B}_x(r) \Delta \mathbf{B}_y(r)) < 2c(r + 1)^n\varepsilon$$
2. Let $m \in \mathbb{N}_1$ and consider some pair of $m$-length sequences in $\mathcal{B}_n$: $(B_1, \dots, B_m)$, $(C_1, \dots, C_m)$ with $B_i, C_i \in \mathcal{B}_n$ for every $i \in \{1, \dots, m\}$. Let $\varepsilon \in (0,\infty)$. If, for each $i \in \{1, \dots, m\}$, we have $\lambda_n(B_i \Delta C_i) < \varepsilon$, then $$\lambda_n\left(\left(\bigcup_{i = 1}^m B_i\right) \Delta \left(\bigcup_{i = 1}^m C_i\right)\right) < m\varepsilon$$
3. Let $m\in \mathbb{N}_1$, let $\varepsilon \in (0,\infty)$, let $x := (x_1, \dots, x_m), y := (y_1, \dots, y_m) \in (\mathbb{R}^n)^m$ and let $r \in (0,\infty)$. If, for every $i \in \{1, \dots, m\}$, $|x_i - y_i|_n < \varepsilon$, we have $$\lambda_n\left(\left(\bigcup_{i = 1}^m \mathbf{B}_{x_i}(r)\right) \Delta \left(\bigcup_{j = 1}^m \mathbf{B}_{y_j}(r)\right)\right) < 2mc(r + 1)^n\varepsilon$$
Proof
1. By the triangle inequality, $\mathbf{B}_y(r) \subseteq \mathbf{B}_x(r + |y - x|_n)$. Hence $\mathbf{B}_y(r) \setminus \mathbf{B}_x(r) \subseteq \mathbf{B}_x(r + |y - x|_n) \setminus \mathbf{B}_x(r)$. Hence $$\lambda_n(\mathbf{B}_y(r) \setminus \mathbf{B}_x(r)) \leq \lambda_n(\mathbf{B}_x(r + |y - x|_n)) - \lambda_n(\mathbf{B}_x(r))$$
By the formula for the volume of an $n$-dimensional ball, \begin{aligned} \lambda_n(\mathbf{B}_x(r + |y - x|_n)) - \lambda_n(\mathbf{B}_x(r)) & = c(r + |y - x|_n)^n - cr^n \\ & = c \sum_{k = 1}^n \binom{n}{k}r^{n - k}|y - x|_n^k \\ & = c |y - x|_n \sum_{k = 1}^n \binom{n}{k}r^{n - k}|y - x|_n^{k - 1} \\ & \leq c \varepsilon \sum_{k = 1}^n\binom{n}{k}r^{n - k} \\ & < c \varepsilon (r + 1)^n \end{aligned}
By an analogous argument, $\lambda_n(\mathbf{B}_x(r) \setminus \mathbf{B}_y(r)) < c \varepsilon (r + 1)^n$.
2. By properties of the symmetric difference, $$\left(\bigcup_{i = 1}^m B_i\right) \Delta \left(\bigcup_{i = 1}^m C_i\right) \subseteq \bigcup_{i = 1}^m \left(B_i \Delta C_i\right)$$
3. The result follows from the first two parts.
Q.E.D.
Definition 5
For every $m \in \mathbb{N}_1$, every $t \in (0,\infty)$ and every $r \in (0,\infty)$,
1. Define $e_{m,r} : (\mathbb{R}^n)^m \rightarrow 2^{\mathbb{R}^n}$ and $E_{m,r} : (\mathbb{R}^n)^m \rightarrow [0,\infty)$ as follows \begin{align} e_{m,r}(x_1, \dots, x_m) & := \bigcup_{i = 1}^m\overline{\mathbf{B}_{x_i}(r)} \\ E_{m,r}(x_1, \dots, x_m) & := \lambda_n(e_m(x_1, \dots, x_m)) \end{align}
2. Define $f_{m, r, t} : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $F_{m, r, t} : \Omega \rightarrow [0,\infty)$ as follows \begin{align} f_{m, r, t}(\omega) & := e_{m,r}(W(\omega, 0), W(\omega, \frac{t}{m}), W(\omega, 2\frac{t}{m}), \dots, W(\omega, t)) \\ F_{m, r, t}(\omega) & := \lambda_n(f_{m, r, t}(\omega)) \end{align}
3. Define $g_t : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $G_t : \Omega \rightarrow [0,\infty]$ as follows \begin{align} g_t(\omega) & := \{W(\omega, s) :\mid s \in [0,t]\} \\ G_t(\omega) & := \lambda_n^*(g_t(\omega)) \end{align}
4. Define the functions $g^* : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $G^* : \Omega \rightarrow [0,\infty]$ as follows \begin{aligned} g^*(\omega) & := \{W(\omega, t) :\mid t \in [0,\infty)\} \\ G^*(\omega) & := \lambda_n^*(g^*(\omega)) \end{aligned}
5. Define $h_{r, t} : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $H_{r, t} : \Omega \rightarrow [0,\infty]$ as follows \begin{align} h_{r, t}(\omega) & := \bigcup_{s \in [0, t]} \overline{\mathbf{B}_{W(\omega, s)}(r)} \\ H_{r, t}(\omega) & := \lambda_n^*(h_{r, t}(\omega)) \end{align}
Lemma 6
For every $m \in \mathbb{N}_1$, every $r \in (0,\infty)$, every $t \in (0,\infty)$, every sequence $r_1, r_2, \dots \in (0,\infty)$ such that $\lim_{k \rightarrow \infty} r_k = 0$ and every $\omega \in \Omega$,
1. $h_{r, t}(\omega)$ is compact (and therefore $\in \mathcal{B}_n$ and $\lambda_n^*(h_{r, t}(\omega)) = \lambda_n(h_{r, t}(\omega))$).
2. For every $r' \in [r,\infty)$, we have \begin{aligned} h_{r, t}(\omega) & \subseteq h_{r', t}(\omega) \\ H_{r, t}(\omega) & \leq H_{r', t}(\omega) \end{aligned}
3. We have $$g_t(\omega) = \bigcap_{k = 1}^\infty h_{r_k, t}(\omega)$$ (and therefore $g_t(\omega) \in \mathcal{B}_n$ and $\lambda_n^*(g_t(\omega)) = \lambda_n(g_t(\omega)$).
4. We have $$G_t(\omega) = \lim_{k \rightarrow \infty} H_{r_k, t}(\omega)$$
5. We have \begin{align} f_{m, r, t}(\omega) & \subseteq h_{r, t}(\omega) \\ F_{m, r, t}(\omega) & \leq H_{r, t}(\omega) \end{align}
6. $E_{m, r}$ is $\mathcal{O}_{n; m}/\mathcal{O}_1$-continuous (and hence $(\otimes_{i = 1}^m\mathcal{B}_n)/\mathcal{B}_1$-measurable).
7. $F_{m, r, t}$ is $\mathcal{F}/\mathcal{B}_1$-measurable.
8. $g^*(\omega) \in \mathcal{B}_n$ (and therefore $\lambda_n^*(g^*(\omega)) = \lambda_n(g^*(\omega))$.
Proof
1. Define $d : \mathbb{R}^n \rightarrow [0,\infty)$ as follows $$d(x) := \inf \{|x - y|_n :\mid y \in g_t(\omega)\}$$ Then $h_{r, t}(\omega) = d^{-1}([0,r])$. Hence, by the fact that $d$ is $\mathcal{O}_n / \mathcal{O}_1$-continuous ([M2] p. 175), $h_{r, t}(\omega)$ is closed. Since $[0,t]$ is compact, so is $g_t(\omega)$. In particular, $g_t(\omega)$ is bounded. Let $R \in (0,\infty)$ be such that $g_t(\omega) \subseteq \overline{\mathbf{B}_0(R)}$. Then $h_{r, t}(\omega) \subseteq \overline{\mathbf{B}_0(R + r)}$. In particular, $h_{r, t}(\omega)$ is bounded. $h_{r, t}(\omega)$ is closed and bounded, hence compact.
2. Immediate.
3. It is evident that $$g_t(\omega) \subseteq \bigcap_{k = 1}^\infty h_{r_k, t}(\omega)$$ I will prove the converse containment. Let $x \in \bigcap_{k = 1}^\infty h_{r_k, t}(\omega)$. For every $k \in \mathbb{N}_1$, let $y_k \in g_t(\omega)$ be such that $|y_k - x|_n \leq r_k$. Since $g_t(\omega)$ is compact, there is, by the Bolzano-Weierstrass theorem, a convergent subsequence $(y_{k_1}, y_{k_2}, \dots)$. Set $y := \lim_{i \rightarrow \infty} y_{k_i}$. Since $g_t(\omega)$ is closed, $y \in g_t(\omega)$. I will show that $x = y$.
Let $\varepsilon \in (0,\infty)$ and let $i \in \mathbb{N}_1$ be such that $r_{k_i} < \varepsilon / 2$ and $|y_{k_i} - y|_n < \varepsilon / 2$. We have, by the triangle inequality, $$|y - x|_n \leq |y - y_{k_i}|_n + |y_{k_i} - x|_n < \varepsilon$$ Since $\varepsilon$ was chosen arbitrarily in $(0,\infty)$, $|y - x|_n = 0$.
4. Choose a strictly decreasing subsequence $(r_{k_1}, r_{k_2}, \dots)$. Then by 6.2 and 6.3, $$g_t(\omega) = \bigcap_{i = 1}^\infty h_{r_{k_i}, t}(\omega)$$ Hence, by the continuity of measures, and by the fact (6.1) that $h_{r_{k_1}, t}$ is bounded and therefore of finite Lebesgue measure, $$G_t(\omega) = \lim_{i \rightarrow \infty} H_{r_{k_i}, t}(\omega)$$ But by 6.2, $\lim_{i \rightarrow \infty} H_{r_{k_i}, t}(\omega) = \lim_{k \rightarrow \infty} H_{r_k, t}(\omega)$.
5. Immediate.
6. Fix some $x_1, \dots, x_m \in \mathbb{R}^n$. For every $y_1, \dots, y_m \in \mathbb{R}^n$, we have $$|E_{m, r}(y_1, \dots, y_m) - E_{m, r}(x_1, \dots, x_m)| \leq \lambda_n(e_{m, r}(y_1, \dots, y_m) \Delta e_{m, r}(y_1, \dots, y_m))$$ Let $\varepsilon \in (0,\infty)$. Define $$\delta := \frac{\varepsilon}{2 m c (r + 1)^n}$$ If $|y_i - x_i|_n < \delta$ for all $i \in \{1, \dots, m\}$, then, by lemma 4.3, $$\lambda_n(e_{m, r}(y_1, \dots, y_m) \Delta e_{m, r}(y_1, \dots, y_m)) < \varepsilon$$
7. Observe that, for all $\omega \in \Omega$, $$F_{m, r, t}(\omega) = E_{m, r}(W(\omega, 0), W(\omega, \frac{t}{m}), \dots, W(\omega, t))$$ The result now follows by 6.6.
8. The result follows from 6.3 by observing that $$g^*(\omega) = \bigcup_{k = 1}^\infty g_k(\omega)$$
Q.E.D.
Definition 7
1. Define $u : (n - 2, \infty) \rightarrow (0,\infty)$ as follows $$u(s) := \frac{1}{\sqrt{\pi}} \frac{s}{s - (n - 2)} \exp\left(-\frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right)\right)$$ (for the rationale - see the next lemma.)
2. For every $a, b \in [0, \infty)$ with $a < b$, define $d_{a, b}: \Omega \rightarrow [0,\infty]$ as follows $$d_{a, b}(\omega) := \max \{\left|W(\omega, s) - W(\omega, a)\right|_n :\mid s \in [a, b]\}$$
3. For every $m \in \mathbb{N}_1$ and every $t \in (0,\infty)$, define $D_{m, t} : \Omega \rightarrow [0,\infty)$ as follows $$D_{m, t}(\omega) := \max(d_{0, \frac{1}{m}t}(\omega), d_{\frac{1}{m}t, \frac{2}{m}t}(\omega), \dots, d_{\frac{m - 1}{m}t, t}(\omega))$$
Lemma 8
1. $\lim_{s \rightarrow \infty} s u(s) = 0$
2. Let $a, b \in [0, \infty)$ with $a < b$.
a) $d_{a,b}$ is $\mathcal{F}/\mathcal{B}_1$ - measurable (and, therefore, so is $D_{m, t}$ for all $m \in \mathbb{N}_1$, $t \in (0, \infty)$).
b) $d_{a, b} \sim d_{0, b - a}$,
c) For every $r \in (0,\infty)$, $$\frac{r^2}{b} > n - 2 \implies P(d_{0, b} > r) \leq u\left(\frac{r^2}{b}\right)$$
3. Let $r \in (0,\infty)$ and $t \in (0,\infty)$.
a) For every $m \in \mathbb{N}_1$, $$\{D_{m , t} \leq r \} \subseteq \{g_t \subseteq f_{m, r, t}\} = \{G_t \leq F_{m, r, t}\}$$ b) There exists some $M \in \mathbb{N}_1$, such that, for all $m \in \{M, M + 1, \dots\}$, $$P(D_{m, t} > r) \leq m u\left(\frac{r^2}{t} m\right)$$
Proof
1. Since, for every $s \in (0,\infty)$, $$\frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right) = \frac{1}{2} \frac{s}{n} \left(n - (n - 2) \frac{\ln(s / n)}{s / n} + \frac{\ln(n) - n}{s / n} \right)$$
we have
$$\lim_{s \rightarrow \infty} \frac{\frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right)}{\frac{s}{2\sqrt{\pi}}} = \sqrt{\pi} > 1$$
Hence $$\limsup_{s \rightarrow \infty} \frac{\exp\left(- \frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right)\right)}{\exp(- \frac{s}{2\sqrt{\pi}})} \leq 1$$
Hence $$\limsup_{s \rightarrow \infty} \frac{s u(s)}{\frac{s}{\sqrt{\pi}} \exp(- \frac{s}{2\sqrt{\pi}})} \leq 1$$
But $$\lim_{s \rightarrow \infty} \frac{s}{\sqrt{\pi}} \exp(- \frac{s}{2\sqrt{\pi}}) = 0$$
Therefore, $\limsup_{s \rightarrow \infty} s u(s) = 0$.
2. a) Let $\omega \in \Omega$. Since $W(\omega)$ is continuous and since the rational numbers are dense in $\mathbb{R}$, we have $$d_{a, b}(\omega) = \sup \{\left|W(\omega, q) - W(\omega, a)\right|_n :\mid q \in [a, b] \cap \mathbb{Q}\}$$ The result now follows from the fact ([K2] theorem 1.92, p. 40) that the supremum of a denumerable family of random variables is a random variable.
b) Define $\varphi: \mathbf{C}_{(0; n)} \rightarrow [0,\infty)$ as follows $$\varphi(f) := \max f([0,t])$$ Firstly, I will show that $\varphi$ is $\mathcal{B}(\mathbf{C}_{(0; n)})/\mathcal{B}_1$-measurable. It suffices to show that for all $M \in \mathbb{R}$, $\{\varphi \leq M\} \in \mathcal{B}(\mathbf{C}_{(0; n)})$. Let $M \in \mathbb{R}$. We have $$\{\varphi \leq M\} = \bigcap_{s \in [0,t]} \{f \in \mathbf{C}_{(0; n)} \mid: f(s) \leq M\} = \bigcap_{q \in [0,t] \cap \mathbb{Q}} \{f \in \mathbf{C}_{(0; n)} \mid: f(q) \leq M\}$$ For every $q \in [0,t] \cap \mathbb{Q}$, $\{f \in \mathbf{C}_{(0; n)} \mid: f(q) \leq M\} \in \mathcal{B}(\mathbf{C}_{(0; n)})$, hence $\{\varphi \leq M\} \in \mathcal{B}(\mathbf{C}_{(0; n)})$.
Now, define the process $V : \Omega \times [0,\infty) \rightarrow \mathbb{R}^n$ as follows: $$V(\omega, t) := W(\omega, a + t) - W(\omega, a)$$ $V$ is a standard, $n$-dimensional Brownian motion ([S] 2.9, p. 12), hence $V \sim W$. Therefore, $$d_{a, b} = \varphi \circ V \sim \varphi \circ W = d_{0, b - a}$$ c) For every $k \in \{1, \dots, n\}$, define $$d_{0, b; k}(\omega) := \max \{\left|W_k(\omega, s) - W_k(\omega, a)\right|_1 :\mid s \in [0, b]\}$$ We have $$d_{0, b} \leq \sqrt{d_{0, b; 1}^2 + \cdots + d_{0, b; n}^2}$$ Hence, \begin{aligned} P(d_{0, b} > r) & = P(d_{0, b}^2 > r^2) & \\ & \leq P(\sum_{k = 1}^n d_{0, b; k}^2 > r^2) & \\ & = P(\sum_{k = 1}^n W_k^2(b) > r^2) & (1) \\ & = \chi^2(n)\left(\left(r^2/b, \infty\right)\right) & (2) \\ & \leq u\left(\frac{r^2}{b}\right) & (3) \end{aligned} where (1) is by Bachelier's maximum process theorem ([K1] Proposition 13.13, p. 256) together with the fact that the components of an $n$-dimensional Brownian motion are independent; $\chi^2(n)$ in (2) denotes the (central) chi-squared distribution with $n$ degrees of freedom; and (3) is from [I2] equation (3.1), p. 341 (see also [I1] Lemma 1, p. 586; a proof can be found there in the appendix).
3. a) Let $\omega \in \{D_{m, r} \leq r\}$ and let $x \in g_t(\omega)$. Let $s \in [0,t]$ be such that $x = W(\omega, s)$. Define $$K := \max \{k \in \{0, 2, \dots, m - 1\} \mid: s \geq \frac{k}{m}\}$$ Since $d_{K, K + 1}(\omega) \leq r$, $g_s(\omega) \in \mathbf{B}_{g_{K / m}(\omega)}(r) \subseteq f_{m, r, t}(\omega)$.
b) Let $m \in \mathbb{N}_1$ be such that $$\frac{r^2}{t} m > n - 2$$ Then \begin{aligned} P(D_{m,t} > r) & \leq \sum_{k = 0}^{m - 1} P(d_{k\frac{t}{m}, (k + 1) \frac{t}{m}} > r) & \\ & = m P(d_{0, \frac{t}{m}} > r) & (1)\\ & \leq m u\left(\frac{r^2}{t} m\right) & (2) \end{aligned} where (1) is by 8.2 (b) and (2) is by 8.2 (c). Now set $$M := \lceil\frac{t}{r^2} (n - 2)\rceil + 1$$
Q.E.D.
Theorem 9
For every $t \in (0, \infty)$,
1. $G_t$ is $\mathcal{F}/\mathcal{B}_1$-measurable in the sense that there is an $\mathcal{F}/\mathcal{B}_1$-measurable function $G'_t$ and an event $B \in \mathcal{F}$, with $P(B) = 0$, such that $$\{G'_t \neq G_t\} \subseteq B$$
2. $G^*$ is $\mathcal{F}/\mathcal{B}_1$-measurable (in the same sense as in 9.1).
Proof
1. For every $k \in \mathbb{N}_1$, use 8.1 to choose $I_k \in \mathbb{N}_1$ such that, for all $i \in \{I_k, I_k + 1, \dots\}$, $$\frac{k^{-2}}{t} i \cdot u(\frac{k^{-2}}{t} i) < \frac{k^{-2}}{t} \cdot \frac{1}{k}$$ and use 8.3 (b) to choose $J_k \in \mathbb{N}_1$, such that, for all $j \in \{J_k, J_k + 1, \dots\}$, $$P(D_{j, t} > \frac{1}{k}) \leq j u\left(\frac{k^{-2}}{t} j\right)$$ Set $m_k := \max(I_k, J_k)$, and define $$A_k := \left\{D_{m_k, t} \leq \frac{1}{k}\right\}$$ Then $P(A_k) \geq 1 - \frac{1}{k}$.
For every $k \in \mathbb{N}_1$ we have, by 6.5, \begin{aligned} &F_{m_k, \frac{1}{k}, t} \leq H_{\frac{1}{k}, t} & (1) \end{aligned}
and by 6.2 and 6.4 \begin{aligned} &G_t = \lim_{j \downarrow \infty} H_{\frac{1}{j}, t} & (2) \end{aligned}
and by 8.3 (a), \begin{aligned} &\forall \omega \in A_k, G_t(\omega) \leq F_{m_k, \frac{1}{k}, t}(\omega) & (3) \end{aligned}
The result now follows from (1), (2) and (3) by theorem 1 here.
2. Observe that for every $\omega \in \Omega$, $$g^*(\omega) = \lim_{k \uparrow \infty} g_k(\omega)$$ Therefore, by continuity of measures, $G^* = \lim_{k \rightarrow \infty} G_k$. The result now follows from 9.1.
Q.E.D.
Remark 10
The only snag that prevents this proof from extending to the case $n = 1$, is the requirement that $n \geq 2$ in equation (3.1) of [I2], which is used to justify the upper bound used in 8.2 (c).
To extend to $n = 1$, the following bound on the expression that is tagged with "(1)" in the proof of 8.2 (c) can be used (cf. [M1] lemma 12.9, p. 349): $$1 - \Phi(r) \leq \frac{1}{r} \frac{1}{\sqrt{2\pi}} e^{- r^2 / 2}$$ where $\Phi$ is the c.d.f. of the standard normal distribution.
WORKS CITED
• [I1] Inglot, Tadeusz and Ledwina, Teresa. Asymptotic optimality of a new adaptive test in regression model, Annales de l'Institut Henri Poincaré 42 (2006), pp. 579–590.
• [I2] Inglot, Tadeusz. Inequalities for Quantiles of the Chi-Square Distribution. Probability and Mathematical Statistics, Vol. 30, Fasc. 2 (2010), pp. 339-351
• [K1] Kallenberg, Olav. Foundations of modern probability. 2nd edition. Springer, 2001
• [K2] Klenke, Achim. Probability Theory, A Comprehensive Course. Springer, 2008
• [L] Lévy, Paul M. Le mouvement Brownien plan. American Journal of Mathematics, Vol 62, No. 1 (1940), pp. 487-550
• [M1] Mörters, Peter and Peres, Yuval. Brownian motion. Published by Cambridge University Press, but I used the online version from Yuval Peres's homepage, retrieved 2014-02-18
• [M2] Munkres, James R. Topology. 2nd edition. Prentice Hall, 2000
• [S] Schilling, René L. and Partzsch, Lothar. Brownian Motion, An Introduction to Stochastic Processes. De Gruyter, 2012
• Wow, what an answer. Just for future reference, you may find it useful to draft your response using our sandbox for long answers. It helps the community in that by lowering the number of edits, you free up some real-estate on the front page for other questions. Thanks! – Willie Wong Apr 10 '14 at 7:19
• @WillieWong: Thanks, will do. I did actually use a separate sandbox (stackedit.io) to prepare this answer, which I've been working on for several days, but there are always bugs, omissions and pitfalls that I find after I post my answers. It can't be helped! – Evan Aad Apr 10 '14 at 7:30 | 10,558 | 26,267 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-22 | latest | en | 0.88312 |
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GRE vs. GMAT
Author Message
VP
Joined: 08 Apr 2009
Posts: 1212
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
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01 Mar 2010, 07:57
4
KUDOS
With top schools now accepting both GRE and GMAT. I thought this might be a common question.
GRE and GMAT Score Comparison Tool
Attachment:
GRE_Comparison_Tool.xls [124.5 KiB]
GRE vs GMAT, Which Test to Take?
Intern
Joined: 17 Mar 2010
Posts: 4
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17 Mar 2010, 05:24
GMAT is still more popular between MBA programs as well as Students
Founder
Joined: 04 Dec 2002
Posts: 16886
Location: United States (WA)
GMAT 1: 750 Q49 V42
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17 Mar 2010, 17:47
Sure by far: gre-gets-little-traction-in-bschools-91370.html
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05 May 2010, 07:44
Yes, both have their advantages and disadvantages. On one hand, if you ever decide that you want to go for a master's in an academic field, the GRE is vital. On the other hand, not every business school out there takes a GRE score. However, one part of the article I disagree with is that the GMAT provides more security. The GRE administrators take just as many measures and those of the GMAT to make sure that the person actually taking the test is the person whose name is on the form. Personally, since I don't know exactly where I want to go yet, I plan on taking both.
Intern
Joined: 08 Mar 2010
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06 May 2010, 13:16
1
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Actually, that's a good thing to mention - the fact that not all business schools accept the GRE score. It might not be time well-spent to study and take a gre practice test when working toward a good GMAT score is time better spent.
ahillis wrote:
Yes, both have their advantages and disadvantages. On one hand, if you ever decide that you want to go for a master's in an academic field, the GRE is vital. On the other hand, not every business school out there takes a GRE score.
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Joined: 05 Jan 2010
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06 May 2010, 13:57
joeleitz wrote:
Actually, that's a good thing to mention - the fact that not all business schools accept the GRE score. It might not be time well-spent to study and take a gre practice test when working toward a good GMAT score is time better spent.
This. Plus its unclear how GRE performance is evaluated relative to GMAT performance in the eyes of each admissions committee. Do they take them both at face value? Do they map your GRE Q/V/Total percentiles onto the equivalent GMAT scores? Do they weight the splits differently? Is not taking the GMAT or not sending the GMAT score you received a negative, in their eyes? Is sending both scores a negative? Is it a positive?
It adds another variable to the decision equation and creates more uncertainty for future applicants -- something you might not want during a process that many see as already being a bit of a crapshoot for certain profiles.
That said, I took both because I was a dual degree applicant and some of my non-MBA applications required the GRE.
Manager
Joined: 22 Oct 2009
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GMAT 1: 760 Q49 V44
GPA: 3.88
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03 Jul 2010, 07:33
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I think that the GRE should be taken advantage of by people who struggle with the GMAT quant section. The GRE math section is a cakewalk in comparison
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03 Jul 2010, 17:34
YourDreamTheater wrote:
I think that the GRE should be taken advantage of by people who struggle with the GMAT quant section. The GRE math section is a cakewalk in comparison
Honestly, I didn't find this to be the case, but perhaps that's just me. The GMAT may offer problems that are somewhat more complex, but it also allows test takers 25% more time to solve them. I mean, the material is exactly the same. I suppose if someone prefers the GRE's "Which value is bigger?" to the GMAT's "Which of these things do you need to solve the problem?" he/she might find the GRE more intuitive. But I wouldn't say either is inherently more difficult.
IMO, most MBA applicants should still prepare for the GMAT, unless they have a concrete reason otherwise, such as a plan to apply to a degree program that won't accept it.
Manager
Joined: 22 Oct 2009
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06 Jul 2010, 15:02
coaks wrote:
YourDreamTheater wrote:
I think that the GRE should be taken advantage of by people who struggle with the GMAT quant section. The GRE math section is a cakewalk in comparison
Honestly, I didn't find this to be the case, but perhaps that's just me. The GMAT may offer problems that are somewhat more complex, but it also allows test takers 25% more time to solve them. I mean, the material is exactly the same. I suppose if someone prefers the GRE's "Which value is bigger?" to the GMAT's "Which of these things do you need to solve the problem?" he/she might find the GRE more intuitive. But I wouldn't say either is inherently more difficult.
IMO, most MBA applicants should still prepare for the GMAT, unless they have a concrete reason otherwise, such as a plan to apply to a degree program that won't accept it.
To each their own!
Intern
Joined: 03 Jul 2010
Posts: 8
Schools: Stanford, MIT
WE 1: 1600 GRE
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06 Jul 2010, 15:50
My \$0.02, since this is near and dear to me as a soon-to-be GRE score submitter... The "official" reason that some bschools are taking the GRE is that they want to encourage non-standard applicants (attracted by the lower price of the GRE or the fact that they've already taken it) to apply.
Also, at a Stanford info session I asked about the GMAT vs. GRE and they said "either's fine; we actually think the GRE is harder."
AJ
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09 Jul 2010, 13:26
The GMAT is still preferred, but the GRE is becoming more widely accepted.
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Current Student
Joined: 02 Jul 2010
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14 Jul 2010, 16:02
i took gre
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Intern
Joined: 08 May 2010
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GMAT 1: 750 Q47 V47
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14 Jul 2010, 18:20
I appeared for GRE 5 years ago and got 1460/1600 with 2 weeks prep approx... i'm finding GMAT much much harder.. sometimes I'm wondering whether I should take the GRE again instead of GMAT but i guess its too late
Intern
Joined: 30 Apr 2010
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17 Jul 2010, 04:43
Thanks for the comparision kit ...
Moderator
Status: battlecruiser, operational...
Joined: 25 Apr 2010
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03 Aug 2010, 22:15
GMAT > GRE
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14 Aug 2010, 06:54
Could anyone recommend a similar forum like gmatclub for the GRE. My brother is interested in completing his GRE and is on the lookout for good forums that he can learn from.
Mhasan
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Joined: 06 Nov 2009
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14 Aug 2010, 07:38
Just search in google for gre forum. There are plenty of good gre forums available!
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30 Aug 2010, 10:58
occamsrazor wrote:
I appeared for GRE 5 years ago and got 1460/1600 with 2 weeks prep approx... i'm finding GMAT much much harder.. sometimes I'm wondering whether I should take the GRE again instead of GMAT but i guess its too late
Wow!! 1460 with 2 weeks is amazing
I guess you have an awesome vocabulory. Most candidates take upto 1.5-2 months to study the Barrons wordlists alone.
Intern
Joined: 16 Sep 2011
Posts: 4
Concentration: Entrepreneurship, Strategy
GRE 1: 1450 Q790 V660
WE: Project Management (Consulting)
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16 Sep 2011, 15:59
Is there any way to compare the GRE verbal score to the GMAT verbal score?
Senior Manager
Joined: 25 Jul 2011
Posts: 298
Location: India
Concentration: Strategy, General Management
GMAT 1: 730 Q50 V40
GPA: 3
WE: Operations (Non-Profit and Government)
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22 Sep 2011, 10:05
I'm not sure if there was a tool for comparing the verbal scores, but there definitely isn't one at the moment for the revised GRE.
I'm not sure how relevant it would be, but I've taken a crack at the approximate coversions:
Verbal - GRE Verbal - GMAT 600 37 650 39 700 43 750 47 800 51
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Re: GRE vs. GMAT [#permalink] 22 Sep 2011, 10:05
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Started Apr 12, 2013 | Discussions thread
Like?
Re: Equivalent focal length for MFT lenses In reply to KenBalbari, Apr 15, 2013
KenBalbari wrote:
Detail Man wrote:
KenBalbari wrote:
... there was a recent thread where a less experienced user seemed to be confused about whether getting closer to a subject would change the exposure, with the idea that the inverse square law should apply to the reflected light the same way it does to the lighting source. Now everyone could tell him that this was wrong, and that changing the distance of the camera doesn't change the exposure.
What a piece of work that bloke was. The worst combination ever is dense as well as vitriolic.
Yes, he was trying to BS hs way through, with a "never admit you were wrong" mentality. Basically the world's least successful con man.
When one refers to oneself as a "seasoned professional" after having been run out of town on a rail, it probably helps to have a portfolio of more than wretchedly exposed "bootay shots" to show.
But some of the explanations of why this is so were quite confusing. I read through a couple and was still baffled as to why.
But the simple explanation that clicked for me was that if you move the camera position and keep the focal length the same, the size of the area captured changes so that the total light captured is the same, and if you change the focal length to capture the same scene, the aperture diameter changes to keep the total light (and the exposure) the same. But if you change the distance of your strobe light from the scene, nothing automatically adjusts in the camera, so you have to change your exposure to account for the difference in total light being captured.
Puzzled a bit myself. Found this web-page (linked in the last post on the thread) to be helpful:
http://www.scantips.com/lights/flashbasics1b.html
Regards, DM
I kept wanting to think there was a difference between reflected light and a source light, in that a source light is dispersing, whereas reflected light reaching the camera all tends to be coming off a surface in one direction (and even tends to be polarized for this reason, that only waves hitting the object in a certain way bounce in that direction), which I think is all basically true, but apparently wasn't really that critical here.
In my humble understanding (recently gained from researching that subject a bit), such (reflected) sources represent the integral (in space) of a large number of point-sources (the paths that happen to be reflected in a uniform direction) - similar to the case of a non-point-source radiator.
It's interesting that when the dimension of such a (non-point) source is around 1/5 of the distance from that source, the inverse-square law holds to with around 1% accuracy (or so some state). As ever, though, the area (from a single point-source) decreases with distance, so the intensity (from each individual point-source comprising the whole) remains constant.
It seems that the integral (in space) of a large number of point-sources (might) have different characteristics ? Seems that are still "superpositioned", though. Interesting stuff to try to envision.
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Color scheme? Blue / Yellow | 741 | 3,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2014-52 | longest | en | 0.9645 |
https://classroom.thenational.academy/lessons/reading-analogue-time-to-the-nearest-minute-cdgkjd | 1,701,858,154,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100593.71/warc/CC-MAIN-20231206095331-20231206125331-00688.warc.gz | 202,726,938 | 25,349 | Reading analogue time to the nearest minute
In this lesson, we will learn to read an analogue clock to the nearest minute. We will use vocabulary around 'to' and 'past', and utilise linear time lines to compare times on clocks to number lines.
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.All scales on a clock are red and blue.
1/5
Q2.Which artist made the melting clock sculpture that is in Shanghai, China?
2/5
Q3.How many scales does the clock use?
3/5
Q4.Gemma said we can estimate the time using just the minute hand. Toby said we can estimate the time using just the hour hand. Who is right?
4/5
Q5.Calendars, diaries and birthdays do not have a connection with time.
5/5
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.All scales on a clock are red and blue.
1/5
Q2.Which artist made the melting clock sculpture that is in Shanghai, China?
2/5
Q3.How many scales does the clock use?
3/5
Q4.Gemma said we can estimate the time using just the minute hand. Toby said we can estimate the time using just the hour hand. Who is right?
4/5
Q5.Calendars, diaries and birthdays do not have a connection with time.
5/5
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Lesson summary: Reading analogue time to the nearest minute
Time to move!
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Move around:
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On the spot:
Dance | 556 | 2,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-50 | latest | en | 0.908948 |
https://www.weibull.com/hotwire/issue181/article181.htm | 1,537,930,090,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163146.90/warc/CC-MAIN-20180926022052-20180926042452-00243.warc.gz | 914,202,344 | 5,292 | # Derivations of Failure Rate Equations for Series and Parallel Systems
This article shows the derivations of the system failure rates for series and parallel configurations of constant failure rate components in Lambda Predict.
## Series System Failure Rate Equations
Consider a system consisting of n components in series. For this configuration, the system reliability, Rs, is given by:
where R1, R2, ..., Rn are the values of reliability for the n components. If the failure rates of the components are λ1, λ2,..., λn, then the system reliability is:
Therefore, the system reliability can be expressed in terms of the system failure rate, λS, as:
where and λS is constant. Note that since the component failure rates are constant, the system failure rate is constant as well. In other words, the system failure rate at any mission time is equal to the steady-state failure rate when constant failure rate components are arranged in a series configuration. If the components have identical failure rates, λC, then:
It should be pointed out that if n blocks with non-constant (i.e., time-dependent) failure rates are arranged in a series configuration, then the system failure rate has a similar equation to the one for constant failure rate blocks arranged in series and is given by:
where λS(t) and λi(t) are functions of time. Please see the HotWire article "Failure Rate of a Series System Using Weibull++" for more details about this equation.
## Parallel System Failure Rate Equations
Consider a system with n identical constant failure rate components arranged in a simple parallel configuration. For this case, the system reliability equation is given by:
where RC is the reliability of each component. Substituting the expression for component reliability in terms of the constant component failure rate, λC, yields:
Notice that this equation does not reduce to the form of a simple exponential distribution like for the case of a system of components arranged in series. In other words, the reliability of a system of constant failure rate components arranged in parallel cannot be modeled using a constant system failure rate model.
To find the failure rate of a system of n components in parallel, the relationship between the reliability function, the probability density function and the failure rate is employed. The failure rate is defined as the ratio between the probability density and reliability functions, or:
Because the probability density function can be written in terms of the time derivative of the reliability function, the previous equation becomes:
The reliability of a system of n components in parallel is:
and its time derivative is:
Substituting into the expression for the system failure rate yields:
For constant failure rate components, the system failure rate becomes:
Thus, the failure rate for identical constant failure rate components arranged in parallel is time-dependent. Taking the limit of the system failure rate as t approaches infinity leads to the following expression for the steady-state system failure rate:
Applying L'Hopital's rule one obtains:
So the steady-state failure rate for a system of constant failure rate components in a simple parallel arrangement is the failure rate of a single component. It can be shown that for a k-out-of-n parallel configuration with identical components: | 644 | 3,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-39 | latest | en | 0.92174 |
https://www.cuemath.com/ncert-solutions/exercise-14-4-factorization-class-8-maths/ | 1,623,515,928,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586239.2/warc/CC-MAIN-20210612162957-20210612192957-00300.warc.gz | 671,612,116 | 12,994 | # Exercise 14.4 Factorization- NCERT Solutions Class 8
## Chapter 14 Ex.14.4 Question 1
Find and correct the errors in the statement:
$4(x - 5) = 4x - 5$
### Solution
What is known?
Incorrect mathematical statement.
What is unknown?
Correct mathematical statement.
Reasoning:
Solve L.H.S.
Steps:
${\rm{S}} = 4(x - 5) \ne \rm R.H.S.$
The correct statement is $$4(x - 5) = 4x - 20$$
## Chapter 14 Ex.14.4 Question 2
Find and correct the errors in the statement:
$x(3x + 2) = 3{x^2} + 2$
### Solution
What is known?
Incorrect mathematical statement.
What is unknown?
Correct mathematical statement.
Reasoning:
Solve L.H.S.
Steps:
\begin{align} {\rm{L.H.S}} &= x(3x + 2)\\&= 3{x^2} + 2x\\ {\rm{L.H.S}} &\ne {\rm{R.H.S}} \end{align}
The correct statement is $$x(3x + 2) = 3{x^2} + 2x$$
Instant doubt clearing with Cuemath Advanced Math Program | 293 | 865 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-25 | latest | en | 0.679413 |
http://forum.woodenboat.com/archive/index.php/t-5822.html?s=25b7443f0cd382c89f8af375f9dc5ee8 | 1,563,670,682,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526799.4/warc/CC-MAIN-20190720235054-20190721021054-00342.warc.gz | 58,627,126 | 3,903 | PDA
View Full Version : How lumber is priced?
03-17-2002, 02:27 AM
Scratching my head after looking a little more closely at the reciept I just recieved for some wood I recently ordered. 8/4" x 6"UP x 12'UP = 39 board feet?!? Am I missing something here? Isn't a "board foot" usually 4/4" x 12" x 1'? So my order should have been more like 12 board feet, no? (2 x .5 x 12?) What do the letters "UP" signify? Uncut product perhaps? I had this order re-sawn for me, and it didn't dawn on me when I picked it up to do the math to figure out just how big a hunk of wood it must have originally come from. I WAS a bit taken aback by the total however! Is it possible they might have made such a basic mistake, or am I just not reading this right? Even waste from resawing couldn't acount for that big a discrepency could it? :confused:
[ 03-17-2002, 02:41 AM: Message edited by: Art Read ]
Roger Stouff
03-17-2002, 07:40 AM
Sounds like 12 bf to me. Dunno what they did.
Here's a neat tool I found:
Board foot calculator (http://www.woodbin.com/calcs/tabulator.htm)
htom
03-17-2002, 09:30 AM
Looks like 12 bf to me, too.
If they'd cut down a 10/4 x 8" x 12', I could understand a bill for 20 bf, but I wouldn't like it.
Maybe they price resawing by multiplying the actual board feet x some factor? Maybe they just made a mistake, billed you for a plank on someone else's order?
Don Maurer
03-17-2002, 11:04 AM
Was the invoice for a single piece or several? It sounds like the price per board foot depends on thickness, width and length. I would expect from the bill that you got 39 bd ft from rough stock of 8/4 x 6"+ x 12'+. Roughly 3 pieces 12' long or equivalent.
NormMessinger
03-17-2002, 11:49 AM
Sounds like 12 bd.ft. to me but only the guy that made up the invoice can tell you for sure. Please let us know what he says.
--Norm
Bill Perkins
03-17-2002, 12:05 PM
I think UP means UnPlaned after the resaw .The bill makes no sence , unless they took a slice off a 6 by 6 and charged you for the whole thing .
RGM
03-17-2002, 09:47 PM
So Art, did you get charged for the fall-off from the re-saw (that you didn't receive) in addition to a handling and re-saw fee that was not itemized separately on your invoice? If I could make one guess as to who you purchased this stock from I bet it would be .... , I better not say here. Reach me offline, inquiring minds want to know.
03-18-2002, 12:12 AM
RGM... Based on our phone conversation about plywood a while back, I'd say you've guessed correctly about the supplier. I'm gonna call 'em tommorow and see what's up, so I won't mention any names either until I hear what they have to say. I was, in fact, charged a seperate re-saw fee, and it's on the invoice. I expected that. But I think the board foot amount must be just a ""clerical" error. I had called for priceing on a "ballparked" amount before actually measuring for it on the boat. A couple hours later, after shopping around a bit and checking my measurements, I called back and placed the order, but for about half the amount I'd had quoted to me earlier. The guy who took the order was still refering to his old notes when he wrote it up. I think he must have just plugged the earlier "board foot" amount in while entering the correct information in the "cut" order. It's the only thing that makes sense to me. The numbers would be about right if it had been the original amount. Still, it's hard to imagine them not catching it... (Harder still to imagine ME not "catching" it!) The amount I actually received seems just about right. (plus the re-saw fall off, which they did include...) The 39 board feet would have been WAY too much. And I just can't see that the waste could have possibly amounted to DOUBLE the yeild?
[ 03-18-2002, 12:51 AM: Message edited by: Art Read ]
03-21-2002, 12:57 PM
Just got a call back from the guy that put my order together. "Somebody" goofed. Got a nice apology and a credit for 14 board feet's worth on my card. This will teach me to look at the invoices more carefully from now on!
Paul Scheuer
03-21-2002, 06:55 PM
Art: 39 - 14 = 12 ? What am I missing ? | 1,151 | 4,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-30 | latest | en | 0.945348 |
https://www.physicsforums.com/threads/symmetry-hints-need-please.98827/ | 1,519,079,609,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812841.74/warc/CC-MAIN-20180219211247-20180219231247-00480.warc.gz | 925,149,795 | 14,652 | # Symmetry - hints need please
1. Nov 7, 2005
### Natasha1
Sorry I am not sure if I should put this thread here.. looks like I am going to be told off by the boss
Anyway, here is my question in a triangle ABC, AB = AC and D is a point on AC such that AD = DB = BC. Find the size of the angle BAD? Find the angles of triangle ABC?
Just need a few hints to get this problem started please?
2. Nov 7, 2005
### Fermat
You triangle, ABC, involves three related/interdepandent similar triangles.
Draw out the triangle(s). Try doing it to scale (i.e. try making the triangles look similar).
Label the angles. It should fall out pretty easily.
3. Nov 7, 2005
### Natasha1
I have ADB being isoceles where AD=DB but from there how do you mesure angle A? I know that A and B will be the same in that triangle but how do u get a value?
4. Nov 8, 2005
### Fermat
There are three isosceles triangles. ABC and ADB are two of them. You have to find the third one before you can solve for the angles.
One property of isosceles triangles is that, not only are two sides of equal length, but the angles oppposite them are also equal.
Label all the (equal) angles and it will fall out. | 314 | 1,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-09 | longest | en | 0.937078 |
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Topic: Jarque-Bera statistics guessing Normal or Uniform?
Replies: 3 Last Post: Apr 18, 2013 6:36 AM
Messages: [ Previous | Next ]
Luis A. Afonso Posts: 4,758 From: LIsbon (Portugal) Registered: 2/16/05
Jarque-Bera statistics guessing Normal or Uniform?
Posted: Apr 13, 2013 10:13 AM
Jarque-Bera statistics guessing Normal or Uniform?
The Classical J.Neyman-E. Pearson NHST scheme to decide if a Null Hypothesis should be rejected can be used in order to choose what the preferential distribution to fit a data-problem is. In fact, allied to systematic simulation concerning the Null, H0, Normal Population, and Alternative Hypotheses, H1, Uniform Population, and the sample response towards a GoF test concerning the Null and an adequate Population parameters we can easily achieve our goal.
Here it is intended, to illustrate the progressive ability to discriminate Uniformity from Normality by the Jarque-Bera (ALM) algorithm in conjunction with the Skewness and Excess Kurtosis indexes when sample sizes are let to vary from 50 to 100 (400?000 samples each time). No specific data-sample is used simulated instead does. So, the test will inform how sizes became untenable towards Normality against Uniformity when we compare the observed triple outputs/responses JB, S, and k, codified by [000] to [111], 1 indicating significant result, 0 not.
The proposed Method
Firstly of all the 5% critical values:
____________JBcrit(5%)____________1%______
____50______6.55__6.55_______16.63__16.62__
____60______6.41__6.41_______16.23__16.19__
____70______6.41__6.40_______15.68__15.68__
____80______6.38__6.39_______15.31__15.32__
____90______6.34__6.33_______14.96__14.92__
___100______6.36__6.31_______14.71__14.73__
and
________________5%_____________1%_______
____________Ucrit__Vcrit___
___50_______3.92___3.44______7.39___11.02__
___60_______3.91___3.44______7.30___10.81__
___70_______3.90___3.45______7.25___10.52__
___80_______3.91___3.47______7.19___10.44__
___90_______3.90___3.48______7.12___10.15__
__100_______3.90___3.52______7.03____9.99__
Finally the *ouputs* using JBcrit, Ucrit,Vcrit (5% level)
After carefully consideering vantages and drawbacks to obtain more precise critical values I opt to choose the following ones, taking into account that semi-quantitative goal of the present account is not extremely demanding.
JBcrit= 6.55, 6.41, 6.40, 6.39, 6.35, 6.35
Ucrit = 3.91______________________
Vcrit = 3.44, 3.44, 3.44, 3.47, 3.47, 3.52
______________________________________
Table__ Jarque-Bera test outputs, ALM
__Normal (Gaussian)
__Uniform
__[000]__[001]__[010]___[011]__[100]__[101]__[110]__[111]_
__n=50____
__0.921__0.012__0.017__0.000__0.000__0.016__0.011__0.022_
__0.629__0.369__0.002__0.000__0.000__0.000__0.000__0.000_
__n=60____
__0.919__0.013__0.018__0.000__0.001__0.018__0.012__0.020_
__0.406__0.585__0.002__0.000__0.000__0.007__0.000__0.000_
__n=70____
__0.918__0.013__0.019__0.000__0.001__0.018__0.013__0.019_
__0.228__0.722__0.001__0.000__0.000__0.048__0.001__0.000_
__n=80____
__0.917__0.014__0.019__0.000__0.001__0.019__0.013__0.018_
__0.119__0.721__0.001__0.000__0.001__0.157__0.001__0.000_
__n=90____
__0.916__0.015__0.019__0.000__0.001__0.019__0.013__0.019_
__0.055__0.601__0.000__0.000__0.001__0.341__0.001__0.000_
Supposing that you had found [001] or [101], which is, of course, very likely to occur with H1 true, then you decide to reject H0 because p= 0.601+ 0.341 = 94.2% if. The decision has the risk 0.015+ 0.019 = 3.4% to fail.
On contrary if you got [000] the H1 rejection has only 5.5% probability, and to retain H0, p= 91.6%, is advisable.
__n=100____
__0.915__0.015__0.020__0.000__0.001¬__0.019__0.014__0.016_
__0.025__0.436__0.000__0.000__0.001__0.537__0.001__0.000_
__p = 0.436 + 0.537 = 97.3% if H1=true and get [001] or [101]. Contrasting with p([001]or[101] | H0 true) = 3.4%.
Luis A. Afonso
Date Subject Author
4/13/13 Luis A. Afonso
4/17/13 Luis A. Afonso
4/17/13 Luis A. Afonso
4/18/13 Luis A. Afonso | 1,510 | 4,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-17 | latest | en | 0.770711 |
https://rd.springer.com/chapter/10.1007/978-1-4757-2878-1_13 | 1,524,348,621,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945448.38/warc/CC-MAIN-20180421203546-20180421223546-00239.warc.gz | 681,364,889 | 16,809 | # Geometrical solution of weighted Fermat problem about triangles
• Mario Martelli
Chapter
Part of the Applied Optimization book series (APOP, volume 13)
## Abstract
The paper considers weighted generalization of a classic Fermat problem about triangles: given A, B, C vertices of a triangle and three real positive numbers $$\bar a,\bar b,\bar c,$$ find in plane a point P minimizing the sum of its distances to A, B, C, multiplied by $$\bar a,\bar b,\bar c,$$ respectively. It is shown by simple geometrical methods that, if $$\bar a,\bar b,\bar c,$$ satisfy the triangle inequalities and further conditions also involving the angles of the triangle ABC, then there is one and only one minimum interior point of the triangle and a construction is supplied which enables us to find the minimum point by “ruler and compasses”. In all other cases, one identified vertex of the triangle is the minimum point.
## Keywords
Triangle Inequality Real Positive Number Minimum Point Orthogonal Axis Weber Problem
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
## References
1. 1.
O. Bottema, R. Z. Djordjevié, R. R. Janié, D. S. Mitrinovié, P. M. Vasie, “Geometric Inequalities”, Wolters-Noordhoff Publishing Groningen, pp. 128129, 1968.Google Scholar
2. 2.
F. Heinen, “Progr. Cleve”, pp. 18–19, 1834.Google Scholar
3. 3.
H. Idrissi, O. Lefebvre, C. Michelot, “A primal dual algorithm for a constrained Fermat-Weber problem involving mixed norms”, Operations Research 22, pp. 313–330, 1988.
4. 4.
H. W. Kuhn, “A note on Fermat’s problem”, Mathematical Programming 4, pp. 98–107, 1973.
5. 5.
J. Krarup, S. Vajda, “On Torricelli geometrical solution to a problem of Fermat”, to appear in Special Edition of Journal of Mathematics Applied in Business and Industry: Duality: In celebration of work of S. Vajda; S. Powell and H. P. Williams eds., 1996.Google Scholar
6. 6.
G. Loria, “Storia delle Matematiche”, vol. 1, U. Hoepli, pp. 159–163, Milano, 1929.Google Scholar
7. 7.
T. Simpson, “The Doctrine and Application of Fluxions”, London, 1750.Google Scholar
8. 8.
E. Torricelli, “De Maximis et Minimis”, Opere di Evangelista Torricelli, G. Lo-ria and G. Vassura eds., par. 22, Faenza, Italy, 1919.Google Scholar
9. 9.
G. O. Wesolowsky, “The Weber problem: history and perspectives”, Location Science 1, pp. 5–23, 1993.
10. 10.
D. J. White, “An analogue derivation of the dual of the general Fermat problem”, Management science 23, pp. 92–94, 1976. | 744 | 2,554 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-17 | longest | en | 0.784242 |
https://mathematica.stackexchange.com/questions/211805/trying-to-find-a-better-way-to-implement-a-recursive-depth-first-search | 1,713,106,230,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00088.warc.gz | 374,068,762 | 39,594 | # Trying to find a better way to implement a recursive depth first search
I'm trying to implementa recursive depth first search. It works and the result is below:
e =
{1 \[UndirectedEdge] 2, 1 \[UndirectedEdge] 3,
2 \[UndirectedEdge] 4, 2 \[UndirectedEdge] 5,
3 \[UndirectedEdge] 5, 4 \[UndirectedEdge] 5,
4 \[UndirectedEdge] 6, 5 \[UndirectedEdge] 6};
nG = Graph[e, VertexLabels -> Automatic];
Clear[DFS, getVisited];
getVisited[am_?MatrixQ] :=
Module[{length = Length[am[[1]]]},
visited = ConstantArray[False, length];
visited];
DFS[am_?MatrixQ, n_?IntegerQ] :=
Module[{i, l = Length[visited]},
visited[[n]] = True;
Print["Visited: " , n];
For[i = 1, i <= l, ++i,
If[am[[n, i]] == 1 && visited[[i]] == False, DFS[am, i]]]];
4, 2, 1, 3, 5, 6
The problem is in the implementation. As the list of visited vertices should be global, I need to call the function getVisited before calling my main function to firstly get the global array. This operation seems like a redundant one and I'm looking for another more neat approach. If you have some ideas it would be great.
• What about just using visited = ConstantArray[False, VertexCount[nG]] instead? Dec 22, 2019 at 12:09
• Btw. you might want to let the For loop run only over the is in am["AdjacencyLists"][[n]]. And better use a Do loop instead. Dec 22, 2019 at 12:17
• Also the recursive approach has many shortcomings. One hits the $RecursionLimit quite quickly for larger graphs. Dec 22, 2019 at 12:22 ## 1 Answer Depth-first search is already implemented in the built-in function DepthFirstScan, but I assume you don't just want to use it but learn how to implement this algorithm in Mathematica. I would do it as follows: It is convenient to use an adjacency list representation of the graph. I will use IGAdjacencyList from IGraph/M, which works reliably for all sort of graphs. For simple undirected graphs, you may use AssociationThread[VertexList[g], AdjacencyList[g]] as an alternative that does not require IGraph/M. The implementation: dfs[g_?GraphQ, v0_, f_] := Module[{visited, rec, al = IGAdjacencyList[g]}, If[Not@VertexQ[g, v0], Return[$Failed]]; (* check that the input is valid *)
visited[_] := False;
rec[v_?visited] := Null;
rec[v_] := (visited[v] = True; f[v]; rec /@ al[v]);
rec[v0];
]
g is the input graph, v0 is the starting vertex, and f is a function that will be evaluated for each vertex as they are visited. You may use Sow to collect them in to a list, or Print to just display them.
Demonstration with your graph:
dfs[nG, 4, Print]
4
2
1
3
5
6
Explanation of the function:
• visited is a local symbol which is used to keep track of which vertices have been visited.
• rec is the function that traverses the graph recursively. It takes a vertex as input. For each neighbour of this vertex, 1) it checks if it has already been visited 2) if not, it calls f, marks it as visited, and applies itself to all its neighbours.
Alternatively, you could make visited an association and return its keys at the end.
dfs2[g_?GraphQ, v0_, f_] :=
Module[{visited = <||>, rec, al = IGAdjacencyList[g]},
If[Not@VertexQ[g, v0], Return[\$Failed]];
rec[v_?visited] := Null;
rec[v_] := (AssociateTo[visited, v -> True]; f[v]; rec /@ al[v]);
rec[v0];
Keys[visited]
] | 919 | 3,250 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-18 | latest | en | 0.84396 |
https://www.sapling.com/6076560/calculate-newly-issued-preferred-stock | 1,611,481,234,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00467.warc.gz | 946,551,943 | 120,671 | # How to Calculate The Cost of a Newly Issued Preferred Stock
Determine the cost of newly issued preferred stock.
Calculating how much it will cost a company to issue stock helps that business to determine whether preferred stocks fit into their financial plan. When considering the cost to a company to issue new preferred stock, you should research the company to gather the information needed. Preferred stock differs from common stock since holders of preferred stock usually do not vote on company matters, but their dividends are paid to preferred investors before common shareholders'. Avoid confusing the cost of preferred stock with the price. Cost evaluates the cost to the company to issue the stock as a percentage, while the price refers to the amount of money an investor spends to purchase preferred stock.
## Step 1
Convert the flotation cost percent to a decimal by dividing the number by 100. For example, a 5 percent flotation cost divided by 100 would be: 5/100=0.05
## Step 2
Subtract the decimal of the flotation cost from 1. For the example: 1 – 0.05 = 0.95
## Step 3
Multiply the market price for the preferred stock by one minus the flotation cost. For the example, a market price of \$100 would yield: 100x (0.95) = 95.
## Step 4
Divide the dividend paid by the preferred stock by this number. For the example, a dividend for the stock of \$5 would result in: 5/95 = 0.053
## Step 5
Multiply this result by 100 to find the cost of the newly issued preferred stock as a percent. For the example: 0.053 x 100 = 5.3 percent.
### Things You'll Need
• Flotation cost to the company (percent)
• Dividend for the preferred stock (dollars)
• Market price of the preferred stock (dollars)
• Calculator (optional)
references | 405 | 1,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-04 | longest | en | 0.910501 |
https://avenes.pl/stone/18_Fri_2563.html | 1,632,737,507,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00218.warc.gz | 178,324,811 | 7,875 | formula for calculation of capacity of hammer mill
# formula for calculation of capacity of hammer mill
##### Formula For Calculation Of Capacity Of Hammer Mill
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##### formula for calculation of capacity of hammer mill
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##### formula for calculation of capacity of hammer mill
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##### formula for calculation of capacity of hammer mill
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##### formula for calculation of capacity of hammer mill
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##### formula for calculation of capacity of hammer mill
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##### formula for calculation of capacity of hammer mill
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##### Formula For Calculation Of Capacity Of Hammer Mill
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##### formula for calculation of capacity of hammer mill
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##### formula for calculation of capacity of hammer mill
2013-6-21 formula for calculation of capacity of hammer mill. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for anyel size-reduction requirements including quarry, aggregate, and different kinds of minerals.
##### formula for calculation of capacity of hammer mill
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##### formula for calculation of capacity of hammer mill
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##### formula for calculation of capacity of hammer mill
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##### Formula For Calculation Of Capacity Of Hammer Mill
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##### Formula For Calculation Of Capacity Of Hammer Mill
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##### Formula For Calculation Of Capacity Of Hammer Mill
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##### formula for calculation of capacity of hammer mill -
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##### Formula For Calculation Of Capacity Of Hammer Mill
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##### Formula For Calculation Of Capacity Of Hammer Mill
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##### formula for calculation of capacity of hammer mill
2013-6-21 formula for calculation of capacity of hammer mill. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for anyel size-reduction requirements including quarry, aggregate, and different kinds of minerals.
##### formula for calculation of capacity of hammer mill
hammer mill capacity calculator Hemming Way. hammer mill tip speed calculation gold ore mainland Formula for calculation of capacity of hammer millammer mills hammermills used in the grinding of animal, pet, livestock, tip speed is the speed of the hammer at it s tip or edge furthest away from the rotor, and is calculated by hammer mill formula when the heat Get Price | 2,281 | 11,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-39 | longest | en | 0.758487 |
https://stats.stackexchange.com/questions/197698/why-does-power-other-conditions-equal-decrease-as-the-odds-ratio-increases-abo | 1,718,890,181,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00564.warc.gz | 478,801,746 | 39,881 | # Why does power (other conditions equal) decrease as the odds ratio increases above 2?
I am simulating a logistic regression setting (P(Y=1|x). Estimated proportion of patients with a poor feature at time zero is 40%. I am generating 1000 replications per scenario, sample size ranging from 100 to 200 by 10, for odds ratios ranging from 1.5 to 2.5 by 0.1. I generate x (years) from Uniform~(2,6). Relative power increases appropriately for odds ratios from 1.5 to 2.0. However, relative power decreases for odds ratios greater than 2.0. Does anyone have an explanation for this.
data t;
b0=-0.405;enter code here
do or=1.5 to 2.5 by 0.1;
b1=log(or);
do ss=100 to 200 by 10;
do rep=1 to 1000;
do i=1 to ss;
*years=6*ranuni(34958487);
years=4*ranuni(34958487)+2;
*years=1*rannor(34958487)+3;
lp=b0 + b1*years;
pi=exp(lp)/(1 + exp(lp));
y=ranbin(45823765,1,pi);
output;
end;
end;
end;
end;
run;
proc sort data=t; by b1 ss rep;run;
ods listing close;
ods select ParameterEstimates(persist);
ods output ParameterEstimates=Estimates;
proc logistic descending data=t;
by b1 ss rep;
model y=years;
run;
ods select all;
ods listing;
data reject; set estimates;
if variable="years";
rej=(probchisq<0.05);
or=exp(b1);
run;
proc sort data=reject; by or ss;run;
options pageno=1 pagesize=max;
proc freq data=reject;
by or;
tables ss*rej/ out=power;
title "Rejection Probability";
run;
data power; set power;
if rej=1;
rejprob=count/1000;
keep or ss rejprob;
run;
This is due to ceiling effects. Look at a plot of the success probabilities in some of your simulated data sets when $\beta_1$ is near the higher end of your range, $\log(2.4)$, vs. the lower end, $\log(1.6)$. Toward the higher end, far more of the predicted probabilities get near upper bound of 1. For example, when $\beta_1 = \log(1.6)$, around $15\%$ of predicted probabilities will be above $.95$; when $\beta_1 = \log(2.4)$, that is more like $50\%$. This dynamic makes it harder to determine the precise effect of the predictor, which is reflected in an increased standard error, and thus a smaller power. To get the expected results, you should fix the marginal probability, $P(Y=1)$, across simulations by adjusting $\beta_0$ appropriately.
• The only thing that comes to mind is that you could calculate $$P(Y=1) = E_X (P(Y=1|X))$$ as a function of $\beta_0, \beta_1$. Given the distribution of $X$, that would equal $$f(\beta_0, \beta_1) = \frac{1}{4} \int_{2}^{6} \frac{1}{1 + e^{-(\beta_0 + \beta_1 x)}} dx$$ I don't think you can do this analytically but, for a fixed value of $\beta_1$ and $P(Y=1)$, you can find the proper value of $\beta_0$ by numerically solving for a root of the equation $$f(\beta_0, \beta_1) - P(Y=1)$$ Commented Feb 22, 2016 at 4:00
• You could do fairly well by setting $P(Y=1)$ at the midpoint of time to a fixed value, which you can do by adjusting the intercept until it's right. Commented May 23 at 21:54 | 894 | 2,907 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-26 | latest | en | 0.789718 |
https://www.tradinformed.com/use-excel-to-backtest-a-trading-strategy-using-an-atr-stop-loss/ | 1,696,090,515,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510697.51/warc/CC-MAIN-20230930145921-20230930175921-00610.warc.gz | 1,109,322,435 | 46,529 | # Use Excel to Backtest a Trading Strategy using an ATR Stop-loss
Last Updated on April 14, 2021 by Mark Ursell
This post continues the series of video articles about how to use Microsoft Excel to backtest trading strategies. In this post I show how to calculate a stop-loss using the ATR and then how to backtest the trading strategy.
For the previous video articles see, An Easy Way to use Excel to Backtest a Trading Strategy – Part 2.
## The Average True Range
Developed by J. Welles Wilder the ATR is very popular with traders. On its own, the ATR can be used to measure market volatility and market range. It is also frequently used in other technical indicators such as the SuperTrend indicator and the ADX.
One of the most popular uses for the ATR was developed by Chuck LeBeau and is referred to as the Chandelier Exit. The chandelier exit sets the stop-loss distance as a multiple of the ATR. The ATR reacts to market conditions so when things are calm, the stop-loss will be relatively close and when things are volatile, the stop-loss will be further away. https://www.youtube.com/embed/a7pNSJL2YEU
## Formulas Used:
H-L = High-Low
H-PC = abs(High-Previous Close)
L-PC = abs(Low-Previous Close)
True Range =max(range)
ATR =average(range)
SL =ATRFactor Max Weekly Drawdown = Low-Previous Close Trading Strategy =IF(F34>G34,IF(N35>M34,((F34-M34)/F34)Q34,(F35/F34)*Q34),Q34) | 363 | 1,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | longest | en | 0.884827 |
http://mathhelpforum.com/new-users/202356-similar-figure-problem-print.html | 1,526,874,459,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863923.6/warc/CC-MAIN-20180521023747-20180521043747-00285.warc.gz | 191,439,436 | 3,628 | # Similar figure problem
• Aug 19th 2012, 06:00 PM
william24
Similar figure problem
Find the unknown in each of the following pairs of similar figures.
Attachment 24550
My solution:
Area of DEF
= 42
Area of ABC = 6
6x4cm2/Area of ABC =4/6x4/6
Is this right? What is the correct answer?
• Aug 19th 2012, 09:13 PM
pickslides
Re: Similar figure problem
How is the area of DEF = 16 \$\displaystyle cm^2\$ ?
• Aug 20th 2012, 01:52 AM
william24
Re: Similar figure problem
Triangle area: base x height/2
base=4 height=6
so, 4x6=24
But, I don't know that is it correct.
• Aug 20th 2012, 02:36 AM
cybertutor
Re: Similar figure problem
Firstly, you need to consider the similar shape problem, before you start to think of the Triangle area.
Similar shapes:
Similar shapes have exactly the same angles, as a shape increases in size, the angles stay the same, but the side lengths increase in the same proportion.
the triangles are similar shapes in this question, so as triangle DEF has 2 lengths shown (4cm and 6cm), and triangle ABC has only one length shown, we can use this to help us find out more information. BC and EF in these triangles represent the same part of each shape, but BC is bigger than EF.
How much bigger though?
Well we need to find out the proportion.... BC/EF= the proportional increase....that's 6cm/4cm = 1.5 times bigger.
Therefore all sides in the ABC triangle will be 1.5 times bigger than the corresponding sides in the DEF triangle.
That means the the 6cm side in the DEF triangle x 1.5 = 9cm.....so this is the length of the vertical dotted line on the ABC triangle.
So now we have a little bit more information...which is actually key to finding the total solution to the triangle ABC's Area.
Your initial area of a triangle calculation wasn't a bad start, but you need to realise that 0.5 x Base x Height is for a RIGHT ANGLED triangle only. ABC is a scalene triangle.
To do this part, you need to consider 2 separate right angled triangles.
I have added another couple of points to the triangle:
"x" is the total length of the base of the triangle
"o" is the corner of the 2 dotted lines
Attachment 24556
Now we have 2 Right Angled triangles to consider:
ABO
ACO
To find ABC ( the scalene triangle), we can subtract the area of ACO from the area of ABO
Ok so here are the individual areas by using 1/2 x Base x Height:
Area(ABO) = 9 times X divided by 2 = 4.5X
Area(ACO) = 9 times (X-6) divided by 2 = 4.5X - 27
Area(ABO) - Area(ACO) = 4.5X - (4.5X - 27)
= 4.5X -4.5X + 27
=27 square cm
Hope this helps | 730 | 2,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-22 | latest | en | 0.921182 |
http://en.wikipedia.org/wiki/Normalisable_wavefunction | 1,427,968,945,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427132827069.83/warc/CC-MAIN-20150323174707-00040-ip-10-168-14-71.ec2.internal.warc.gz | 96,707,098 | 58,188 | # Wave function
(Redirected from Normalisable wavefunction)
Not to be confused with Wave equation.
Comparison of classical and quantum harmonic oscillator conceptions for a single spinless particle. The two processes differ greatly. The classical process (A–B) is represented as the motion of a particle along a trajectory. The quantum process (C–H) has no such trajectory. Rather, it is represented as a wave. Panels (C–F) show four different standing wave solutions of the Schrödinger equation. Panels (G–H) further show two different wave functions that are solutions of the Schrödinger equation but not standing waves.
A wave function in quantum mechanics describes the quantum state of an isolated system of one or more particles. There is one wave function containing all the information about the entire system, not a separate wave function for each particle in the system. Its interpretation is that of a probability amplitude. Quantities associated with measurements, such as the average momentum of a particle, can be derived from the wave function. It is a central entity in quantum mechanics and is important in all modern theories, like quantum field theory incorporating quantum mechanics, while its interpretation may differ. The most common symbols for a wave function are the Greek letters ψ or Ψ (lower-case and capital psi).
For a given system, once a representation corresponding to a maximal set of commuting observables and a suitable coordinate system is chosen, the wave function is a complex-valued function of the system's degrees of freedom corresponding to the chosen representation and coordinate system, continuous as well as discrete. Such a set of observables, by a postulate quantum mechanics, are Hermitian linear operators on the space of states representing a set of physical observables, like position, momentum and spin that can, in principle, be simultaneously measured with arbitrary precision. Wave functions can be added together and multiplied by complex numbers to form new wave functions, and hence are elements of a vector space. This is the superposition principle of quantum mechanics. This vector space is endowed with an inner product such that it is a complete metric topological space with respect to the metric induced by the inner product. In this way the set of wave functions for a system form a function space that is a Hilbert space. The inner product is a measure of the overlap between physical states and is used in the foundational probabilistic interpretation of quantum mechanics, the Born rule, relating transition probabilities to inner products. The actual space depends on the system's degrees of freedom (hence on the chosen representation and coordinate system) and the exact form of the Hamiltonian entering the equation governing the dynamical behavior. In the non-relativistic case, disregarding spin, this is the Schrödinger equation.
The Schrödinger equation determines the allowed wave functions for the system and how they evolve over time. A wave function behaves qualitatively like other waves, such as water waves or waves on a string, because the Schrödinger equation is mathematically a type of wave equation. This explains the name "wave function", and gives rise to wave–particle duality. The wave of the wave function, however, is not a wave in physical space; it is a wave in an abstract mathematical "space", and in this respect it differs fundamentally from water waves or waves on a string.[1][2][3][4][5][6][7]
For a given system, the choice of which relevant degrees of freedom to use are not unique, and correspondingly the domain of the wave function is not unique. It may be taken to be a function of all the position coordinates of the particles over position space, or the momenta of all the particles over momentum space, the two are related by a Fourier transform. These descriptions are the most important, but they are not the only possibilities. Just like in classical mechanics, canonical transformations may be used in the description of a quantum system. Some particles, like electrons and photons, have nonzero spin, and the wave function must include this fundamental property as an intrinsic discrete degree of freedom. In general, for a particle with half-integer spin the wave function is a spinor, for a particle with integer spin the wave function is a tensor. Particles with spin zero are called scalar particles, those with spin 1 vector particles, and more generally for higher integer spin, tensor particles. The terminology derives from how the wave functions transform under a rotation of the coordinate system. No elementary particle with spin 32 or higher is known, except for the hypothesized spin 2 graviton. Other discrete variables can be included, such as isospin. When a system has internal degrees of freedom, the wave function at each point in the continuous degrees of freedom (e.g. a point in space) assigns a complex number for each possible value of the discrete degrees of freedom (e.g. z-component of spin). These values are often displayed in a column matrix (e.g. a 2 × 1 column vector for a non-relativistic electron with spin 12).
In the Copenhagen interpretation, an interpretation of quantum mechanics, the squared modulus of the wave function, |ψ|2, is a real number interpreted as the probability density of measuring a particle as being at a given place at a given time or having a definite momentum, and possibly having definite values for discrete degrees of freedom. The integral of this quantity, over all the system's degrees of freedom, must be 1 in accordance with the probability interpretation, this general requirement a wave function must satisfy is called the normalization condition. Since the wave function is complex valued, only its relative phase and relative magnitude can be measured. Its value does not in isolation tell anything about the magnitudes or directions of measurable observables; one has to apply quantum operators, whose eigenvalues correspond to sets of possible results of measurements, to the wave function ψ and calculate the statistical distributions for measurable quantities.
The unit of measurement for ψ depends on the system, and can be found by dimensional analysis of the normalization condition for the system. For one particle in three dimensions, its units are [length]−3/2, because an integral of |ψ|2 over a region of three-dimensional space is a dimensionless probability.[8]
## Historical background
In 1905 Einstein postulated the proportionality between the frequency of a photon and its energy, E = hf,[9] and in 1916 the corresponding relation between photon momentum and wavelength, λ = h/p.[10] In 1923, De Broglie was the first to suggest that the relation λ = h/p, now called the De Broglie relation, holds for massive particles, the chief clue being Lorentz invariance,[11] and this can be viewed as the starting point for the modern development of quantum mechanics. The equations represent wave–particle duality for both massless and massive particles.
In the 1920s and 1930s, quantum mechanics was developed using calculus and linear algebra. Those who used the techniques of calculus included Louis de Broglie, Erwin Schrödinger, and others, developing "wave mechanics". Those who applied the methods of linear algebra included Werner Heisenberg, Max Born, and others, developing "matrix mechanics". Schrödinger subsequently showed that the two approaches were equivalent.[12]
In 1926, Schrödinger published the famous wave equation now named after him, indeed the Schrödinger equation, based on classical energy conservation using quantum operators and the de Broglie relations such that the solutions of the equation are the wave functions for the quantum system.[13] However, no one was clear on how to interpret it.[14] At first, Schrödinger and others thought that wave functions represent particles that are spread out with most of the particle being where the wave function is large.[15] This was shown to be incompatible with how elastic scattering of a wave packet representing a particle off a target appears; it spreads out in all directions.[16] While a scattered particle may scatter in any direction, it does not break up and take off in all directions. In 1926, Born provided the perspective of probability amplitude.[16][17][18] This relates calculations of quantum mechanics directly to probabilistic experimental observations. It is accepted as part of the Copenhagen interpretation of quantum mechanics. There are many other interpretations of quantum mechanics. In 1927, Hartree and Fock made the first step in an attempt to solve the N-body wave function, and developed the self-consistency cycle: an iterative algorithm to approximate the solution. Now it is also known as the Hartree–Fock method.[19] The Slater determinant and permanent (of a matrix) was part of the method, provided by John C. Slater.
Schrödinger did encounter an equation for the wave function that satisfied relativistic energy conservation before he published the non-relativistic one, but discarded it as it predicted negative probabilities and negative energies. In 1927, Klein, Gordon and Fock also found it, but incorporated the electromagnetic interaction and proved that it was Lorentz invariant. De Broglie also arrived at the same equation in 1928. This relativistic wave equation is now most commonly known as the Klein–Gordon equation.[20]
In 1927, Pauli phenomenologically found a non-relativistic equation to describe spin-1/2 particles in electromagnetic fields, now called the Pauli equation.[21] Pauli found the wave function was not described by a single complex function of space and time, but needed two complex numbers, which respectively correspond to the spin +1/2 and −1/2 states of the fermion. Soon after in 1928, Dirac found an equation from the first successful unification of special relativity and quantum mechanics applied to the electron, now called the Dirac equation. In this, the wave function is a spinor represented by four complex-valued components:[19] two for the electron and two for the electron's antiparticle, the positron. In the non-relativistic limit, the Dirac wave function resembles the Pauli wave function for the electron. Later, other relativistic wave equations were found.
### Wave functions and wave equations in modern theories
All these wave equations are of enduring importance. The Schrödinger equation and the Pauli equation are under many circumstances excellent approximations of the relativistic variants. They are considerably easier to solve in practical problems than the relativistic equations. The Klein-Gordon equation and the Dirac equation, while being relativistic, do not represent full reconciliation of quantum mechanics and special relativity. The branch of quantum mechanics where these equations are studied the same way as the Schrödinger equation, often called relativistic quantum mechanics, while very successful, has its limitations (see e.g. Lamb shift) and conceptual problems (see e.g. Dirac sea).
Relativity makes it inevitable that the number of particles in a system is not constant. For full reconciliation, quantum field theory is needed.[22] In this theory, the wave equations and the wave functions has their place, but in a somewhat different guise. The main objects of interest are not the wave functions, but rather operators, so called field operators (or just fields where "operator" is understood) on the Hilbert space of states (to be described next section). It turns out that the original relativistic wave equations and their solutions are still needed to build the Hilbert space. Moreover, the free fields operators, i.e. when interactions are assumed to not exists, turn out to (formally) satisfy the same equation as do the fields (wave functions) in many cases.
Thus the Klein-Gordon equation (spin 0) and the Dirac equation (spin 12) in this guise remain in the theory. Higher spin analogues include the Proca equation (spin 1), Rarita–Schwinger equation (spin 32), and, more generally, the Bargmann–Wigner equations. For massless free fields two examples are the free field Maxwell equation (spin 1) and the free field Einstein equation (spin 2) for the field operators.[23] All of them are essentially a direct consequence of the requirement of Lorentz invariance. Their solutions must transform under Lorentz transformation in a prescribed way, i.e. under a particular representation of the Lorentz group and that together with few other reasonable demands, e.g. the cluster decomposition principle,[24] with implications for causality is enough to fix the equations.
It should finally be emphasized that this is free field equations, interactions are not included. It should also be pointed out that the equations and their solutions are, though needed for the theories, not the central objects of study.
## Wave functions and function spaces
The electron probability density for the first few hydrogen atom electron orbitals shown as cross-sections. These orbitals form an orthonormal basis for the wave function of the electron. Different orbitals are depicted with different scale.
The concept of Function spaces enters naturally in the discussion about wave functions. A function space is a set of functions, usually with some defining requirements on the functions, together with a topology on that set. The latter will sparsely be used here, it is only needed to obtain a precise definition of what it means for a subset of a function space to be closed. A wave function is an element of a function space partly characterized by the following concrete and abstract descriptions.
• The Schrödinger equation is linear. This means that the solutions to it, wave functions, can be added and multiplied by scalars to form a new solution.
• The superposition principle of quantum mechanics. If Ψ and Φ are two states in the abstract space of states of a quantum mechanical system, then aΨ + bΦ is a valid state as well.
The first item says that the set of solutions to the Schrödinger equation is a vector space. The second item says that the set of allowable states is a vector space. This similarity is of course not accidental. Not all properties of the respective spaces have been given so far. There are also a distinctions between the spaces to keep in mind.
• Basic states are characterized by a set of quantum numbers. This is a set of eigenvalues of a maximal set of commuting observables. A choice of such a set may be called a choice of representation. It is a postulate of quantum mechanics that a physically observable quantity of a system, like position, momentum and spin, is represented by a linear Hermitian operator on the state space. The possible outcomes of measurement of the quantity are the eigenvalues of the operator.[15] Maximality refers to that no more algebraically independent linear Hermitian operator can be added to the set that commutes with the ones already present. The physical interpretation is that such a set represents what can – in theory – be simultaneously be measured with arbitrary position at a given time. The set is non-unique. It may for a one-particle system, for example, be position and spin z-projection, (x, Sz), or it may be momentum and spin y-projection, (p, Sy). At a deeper level, most observables, perhaps all, arise as generators of symmetries.[15][25][nb 1]
• Once a representation is chosen, there is still arbitrariness. It remains to choose a coordinate system. This may, for example, correspond to a choice of x, y- and z-axis, or a choice of curvlinear coordinates as exemplified by the spherical coordinates used for the atomic wave functions illustrated below. This final choice also fixes a basis in abstract Hilbert space. The basic states are labeled by the quantum numbers corresponding to the maximal set of commuting observables and an appropriate coordinate system.[nb 2]
• Wave functions corresponding to a state are accordingly not unique. This has been exemplified already with momentum and position space wave functions describing the same abstract state. This non-uniqueness reflects the non-uniqueness in the choice of a maximal set of commuting observables.
• The abstract states are "abstract" only in that an arbitrary choice necessary for a particular explicit description of it is not given. This is the same as saying that no choice of maximal set of commuting observables has been given. This is analogous to a vector space without a specified basis.
• The wave functions of position and momenta, respectively, can be seen as a choice of representation yielding two different, but entirely equivalent, explicit descriptions of the same state for a system with no discrete degrees of freedom.
• Corresponding to the two examples in the first item, to a particular state there corresponds two wave functions, Ψ(x, Sz) and Ψ(p, Sy), both describing the same state. For each choice of maximal commuting sets of observables for the abstract state space, there is a corresponding representation that is associated to a function space of wave functions.
• Each choice of representation should be thought of as specifying a unique function space in which wave functions corresponding to that choice of representation lives. This distinction is best kept, even if one could argue that two such function spaces are mathematically equal, e.g. being the set of square integrable functions. One can then think of the function spaces as two distinct copies of that set.
• Between all these different function spaces and the abstract state space, there are one-to-one correspondences (here disregarding normalization and unobservable phase factors), the common denominator here being a particular abstract state. The relationship between the momentum and position space wave functions, for instance, describing the same state is the Fourier transform.
To make this concrete, in the figure to the right, the 19 sub-images are images of wave functions in position space (their norm squared). The wave functions each represent the abstract state characterized by the triple of quantum numbers (n, l, m), in the lower right of each image. These are the principal quantum number, the orbital angular momentum quantum number and the magnetic quantum number. Together with one spin-projection quantum number of the electron, this is a complete set of observables.
The figure can serve to illustrate some further properties of the function spaces of wave functions.
• In this case, the wave functions are square integrable. One can initially take the function space as the space of square integrable functions, usually denoted L2.
• The displayed functions are solutions to the Schrödinger equation. Obviously, not every function in L2 satisfies the Schrödinger equation for the hydrogen atom. The function space is thus a subspace of L2.
• The displayed functions form part of a basis for the function space. To each triple (n, l, m), there corresponds a basis wave function. If spin is taken into account, there are two basis functions for each triple. The function space thus has a countable basis.
• The basis functions are mutually orthonormal. For this concept to have a meaning, there must exist an inner product. The function space is thus an inner product space. The inner product between two states intuitively measures the "overlap" between the states. The physical interpretation is that the norm squared is proportional to the transition probability between the states. That is.
$P(\Psi \rightarrow \Phi_i) = |(\Psi, \Phi_i)|^2$,
where the i is an index composed of quantum numbers corresponding to a representation and the probabilities are the probabilities of finding the state Ψ in the definite state represented by Φi upon measurement of the physical observables corresponding to the representation, for instance, i could be the quadruple (n, l, m, Sz). This is the Born rule,[16] and is one of the fundamental postulates of quantum mechanics.
These observations encapsulate the essence of the function spaces of which wave functions are elements. Mathematically, this is expressed (in one spatial dimension, disregarding here unimportant issues of normalization) for a particle with no internal degrees of freedom as
$\Psi = I\Psi = \int \Phi_x (\Phi_x, \Psi)dx = \int \Psi(x) \Phi_x dx = \int \Phi_p (\Phi_p, \Psi)dp = \int \Psi(p) \Phi_p dp,$
where Ψ is any "abstract" state, Φx is an eigenfunction of the position operator representing a particle localized at x, (·,·) represents the inner product, Φp is an eigenfunction of the momentum operator representing a particle with precise momentum p, I is the identity operator and the integrals (first and third) represent the completeness of momentum and position eigenstates, Ψ(x) is the coordinate space wave function and Ψ(p) is the wave function in momentum space. In Dirac notation, the above equation reads
$|\Psi\rangle = I|\Psi\rangle = \int |x\rangle \langle x|\Psi\rangle dx = \int \Psi(x) |x\rangle dx = \int |p\rangle \langle p|\Psi\rangle dp = \int \Psi(p) |p\rangle dp.$
The description is not yet complete. There is a further technical requirement on the function space, that of completeness, that allows one to take limits of sequences in the function space, and be ensured that, if the limit exists, it is an element of the function space. A complete inner product space is called a Hilbert space. The property of completeness is crucial in advanced treatments and applications of quantum mechanics. It will not be very important in the subsequent discussion of wave functions, and technical details and links may be found in footnotes like the one that follows.[nb 3] The space L2 is a Hilbert space, with inner product presented later. The function space of the example of the figure is a subspace of L2. A subspace of a Hilbert space is a Hilbert space if it is closed. It is here that the topology of the function space enters into its description.
It is also important to note, in order to avoid confusion, that not all functions to be discussed are elements of some Hilbert space, say L2. The most glaring example is the set of functions e2πipxh. These are solutions of the Schrödinger equation for a free particle, but are not normalizable, hence not in L2. But they are nonetheless fundamental for the description. One can, using them, express functions that are normalizable using wave packets. They are, in a sense to be made precise later, a basis (but not a Hilbert space basis) in which wave functions of interest can be expressed. There is also the artifact "normalization to a delta function" that is frequently employed for notational convenience, see further down. The delta functions themselves aren't square integrable either.
### Physical requirements
Continuity of the wave function and its first spatial derivative (in the x direction, y and z coordinates not shown), at some time t.
The above description of the function space containing the wave functions is mostly mathematically motivated. The function spaces are, due to completeness, very large in a certain sense. Not all functions are realistic descriptions of any physical system. For instance, in the function space L2 one can find the function that takes on the value 0 for all rational numbers and -i for the irrationals in the interval [0, 1]. This is square integrable,[nb 4] but can hardly represent a physical state.
The following constraints on the wave function are sometimes explicitly formulated for the calculations and physical interpretation to make sense:[26][27]
• The wave function must be square integrable. This is motivated by the Copenhagen interpretation of the wave function as a probability amplitude.
• It must everywhere be everywhere continuous and everywhere continuously differentiable. This is motivated by the appearance of the Schrödinger equation.
It is possible to relax these conditions somewhat for special purposes.[nb 5] If these requirements are not met, it is not possible to interpret the wave function as a probability amplitude.[28]
This does not alter the structure of the Hilbert space that these particular wave functions inhabit, but it should be pointed out that the subspace of the square-integrable functions L2, which is a Hilbert space, satisfying the second requirement is not closed in L2, hence not a Hilbert space in itself.[nb 6] The functions that does not meet the requirements are still needed for both technical and practical reasons.[nb 7][nb 8]
## Definition (one spinless particle in 1d)
Travelling waves of a free particle.
The real parts of position wave function Ψ(x) and momentum wave function Φ(p), and corresponding probability densities |Ψ(x)|2 and |Φ(p)|2, for one spin-0 particle in one x or p dimension. The colour opacity of the particles corresponds to the probability density (not the wave function) of finding the particle at position x or momentum p.
For now, consider the simple case of a single particle, without spin, in one spatial dimension. More general cases are discussed below.
### Position-space wave function
The state of such a particle is completely described by its wave function,
$\Psi(x,t)\,,$
where x is position and t is time. This is a complex-valued function of two real variables x and t.
If interpreted as a probability amplitude, the square modulus of the wave function, the positive real number
$\left|\Psi(x, t)\right|^2 = {\Psi(x, t)}^{*}\Psi(x, t) = \rho(x, t),$
is interpreted as the probability density that the particle is at x. The asterisk indicates the complex conjugate. If the particle's position is measured, its location cannot be determined from the wave function, but is described by a probability distribution. The probability that its position x will be in the interval axb is the integral of the density over this interval:
$P_{a\le x\le b} (t) = \int\limits_a^b d x\,|\Psi(x,t)|^2$
where t is the time at which the particle was measured. This leads to the normalization condition:
$\int\limits_{-\infty}^\infty d x \, |\Psi(x,t)|^2 = 1\,,$
because if the particle is measured, there is 100% probability that it will be somewhere.
Since the Schrödinger equation is linear, if any number of wave functions Ψn for n = 1, 2, ... are solutions of the equation, then so is their sum, and their scalar multiples by complex numbers an. Taking scalar multiplication and addition together is known as a linear combination:
$\sum_n a_n \Psi_n(x,t) = a_1 \Psi_1(x,t) + a_2 \Psi_2(x,t) + \cdots$
This is the superposition principle. Multiplying a wave function Ψ by any nonzero constant complex number c to obtain cΨ does not change any information about the quantum system, because c cancels in the Schrödinger equation for cΨ.
### Momentum-space wave function
The particle also has a wave function in momentum space:
$\Phi(p,t)$
where p is the momentum in one dimension, which can be any value from −∞ to +∞, and t is time.
All the previous remarks on superposition, normalization, etc. apply similarly. In particular, if the particle's momentum is measured, the result is not deterministic, but is described by a probability distribution:
$P_{a\le p\le b} (t) = \int\limits_a^b d p \, |\Phi(p,t)|^2 \,,$
and the normalization condition is:
$\int\limits_{-\infty}^{\infty} d p \, \left | \Phi \left ( p, t \right ) \right |^2 = 1\,.$
### Relation between wave functions
The position-space and momentum-space wave functions are Fourier transforms of each other, therefore both contain the same information, and either one alone is sufficient to calculate any property of the particle. As elements of abstract physical Hilbert space, whose elements are the possible states of the system under consideration, they represent the same object, but they are not equal when viewed as square-integrable functions. (A function and its Fourier transform are not equal.) For one dimension,[29]
$\Phi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^\infty d x \, e^{-ipx/\hbar} \Psi(x,t) \quad\rightleftharpoons\quad \Psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^\infty d p \, e^{ipx/\hbar} \Phi(p,t)$
In practice, the position-space wave function is used much more often than the momentum-space wave function. The potential entering the Schrödinger equation determines in which basis the description is easiest. For the harmonic oscillator, x and p enter symmetrically, so there it doesn't matter which description one uses.
## Definitions (other cases)
Traveling waves of two free particles, with two of three dimensions suppressed. Top is position space wave function, bottom is momentum space wave function, with corresponding probability densities.
Following are the general forms of the wave function for systems in higher dimensions and more particles, as well as including other degrees of freedom than position coordinates or momentum components.
The position-space wave function of a single particle in three spatial dimensions is similar to the case of one spatial dimension above:
$\Psi(\mathbf{r},t)$
where r is the position vector in three-dimensional space, and t is time. As always Ψ(r, t) is a complex number, for this case a complex-valued function of four real variables.
If there are many particles, in general there is only one wave function, not a separate wave function for each particle. The fact that one wave function describes many particles is what makes quantum entanglement and the EPR paradox possible. The position-space wave function for N particles is written:[19]
$\Psi(\mathbf{r}_1,\mathbf{r}_2 \cdots \mathbf{r}_N,t)$
where ri is the position of the ith particle in three-dimensional space, and t is time. Altogether, this is a complex-valued function of 3N + 1 real variables.
In quantum mechanics there is a fundamental distinction between identical particles and distinguishable particles. For example, any two electrons are identical and fundamentally indistinguishable from each other; the laws of physics make it impossible to "stamp an identification number" on a certain electron to keep track of it.[29] This translates to a requirement on the wave function for a system of identical particles:
$\Psi \left ( \ldots \mathbf{r}_a, \ldots , \mathbf{r}_b, \ldots \right ) = \pm \Psi \left ( \ldots \mathbf{r}_b, \ldots , \mathbf{r}_a, \ldots \right )$
where the + sign occurs if the particles are all bosons and sign if they are all fermions. In other words, the wave function is either totally symmetric in the positions of bosons, or totally antisymmetric in the positions of fermions.[30] The physical interchange of particles corresponds to mathematically switching arguments in the wave function. The antisymmetry feature of fermionic wave functions leads to the Pauli principle. Generally, bosonic and fermionic symmetry requirements are the manifestation of particle statistics and are present in other quantum state formalisms.
For N distinguishable particles (no two being identical, i.e. no two having the same set of quantum numbers), there is no requirement for the wave function to be either symmetric or antisymmetric.
For a collection of particles, some identical with coordinates r1, r2, ... and others distinguishable x1, x2, ... (not identical with each other, and not identical to the aforementioned identical particles), the wave function is symmetric or antisymmetric in the identical particle coordinates ri only:
$\Psi \left ( \ldots \mathbf{r}_a, \ldots , \mathbf{r}_b, \ldots , \mathbf{x}_1, \mathbf{x}_2, \ldots \right ) = \pm \Psi \left ( \ldots \mathbf{r}_b, \ldots , \mathbf{r}_a, \ldots , \mathbf{x}_1, \mathbf{x}_2, \ldots \right )$
Again, there is no symmetry requirement for the distinguishable particle coordinates xi.
For a particle with spin, the wave function can be written in "position–spin space" as:
$\Psi(\mathbf{r},t,s_z)$
which is a complex-valued function of position r in three-dimensional space, time t, and sz, the spin projection quantum number along the z axis. (The z axis is an arbitrary choice; other axes can be used instead if the wave function is transformed appropriately, see below.) The sz parameter, unlike r and t, is a discrete variable. For example, for a spin-1/2 particle, sz can only be +1/2 or −1/2, and not any other value. (In general, for spin s, sz can be s, s − 1, ... , −s + 1, −s.)
Often, the complex values of the wave function for all the spin numbers are arranged into a column vector, in which there are as many entries in the column vector as there are allowed values of sz. In this case, the spin dependence is placed in indexing the entries and the wave function is a complex vector-valued function of space and time only:
$\Psi(\mathbf{r},t) = \begin{bmatrix} \Psi(\mathbf{r},t,s) \\ \Psi(\mathbf{r},t,s-1) \\ \vdots \\ \Psi(\mathbf{r},t,-(s-1)) \\ \Psi(\mathbf{r},t,-s) \\ \end{bmatrix}$
The wave function for N particles each with spin is the complex-valued function:
$\Psi(\mathbf{r}_1, \mathbf{r}_2 \cdots \mathbf{r}_N, s_{z\,1}, s_{z\,2} \cdots s_{z\,N}, t)$
Concerning the general case of N particles with spin in 3d, if Ψ is interpreted as a probability amplitude, the probability density is:
$\rho\left(\mathbf{r}_1 \cdots \mathbf{r}_N,s_{z\,1}\cdots s_{z\,N},t \right ) = \left | \Psi\left (\mathbf{r}_1 \cdots \mathbf{r}_N,s_{z\,1}\cdots s_{z\,N},t \right ) \right |^2$
and the probability that particle 1 is in region R1 with spin sz1 = m1 and particle 2 is in region R2 with spin sz2 = m2 etc. at time t is the integral of the probability density over these regions and spins:
$P_{\mathbf{r}_1\in R_1,s_{z\,1} = m_1, \ldots, \mathbf{r}_N\in R_N,s_{z\,N} = m_N} (t) = \int\limits_{R_1} d ^3\mathbf{r}_1 \int\limits_{R_2} d ^3\mathbf{r}_2\cdots \int\limits_{R_N} d ^3\mathbf{r}_N \left | \Psi\left (\mathbf{r}_1 \cdots \mathbf{r}_N,m_1\cdots m_N,t \right ) \right |^2$
The multidimensional Fourier transforms of the position or position–spin space wave functions yields momentum or momentum–spin space wave functions.
### Decompositions into products
For systems in time-independent potentials, the wave function can always be written as a function of the degrees of freedom multiplied by a time-dependent phase factor, the form of which is given by the Schrödinger equation. For the case of N particles position-spin space,
$\Psi(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_N,t,s_{z1},s_{z2},\ldots,s_{zN}) = e^{-i Et/\hbar} \psi(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_N,s_{z1},s_{z2},\ldots,s_{zN})\,,$
where E is the energy eigenvalue of the system corresponding to the eigenstate Ψ. Wave functions of this form are called stationary states.
In some situations, the wave function for a particle with spin factors into a product of a space function ψ and a spin function ξ, where each are complex-valued functions, and the time dependence could be placed in either function:
$\Psi(\mathbf{r},t,s_z) = \psi(\mathbf{r},t)\xi(s_z) = \phi(\mathbf{r})\zeta(s_z, t)\,.$
The dynamics of each factor can be studied in isolation. This factorization is always possible when the orbital and spin angular momenta of the particle are separable in the Hamiltonian operator, that is, the Hamiltonian can be split into an orbital term and a spin term.[31] It is not possible for those interactions where an external field or any space-dependent quantity couples to the spin; examples include a particle in a magnetic field, and spin-orbit coupling. For the time-independent case this reduces to
$\Psi(\mathbf{r},t,s_z) = e^{-iEt/\hbar} \psi(\mathbf{r})\xi(s_z) \,,$
where again E is the energy eigenvalue of the system corresponding to the eigenstate Ψ. This extends to the case of N particles:
$\Psi(\mathbf{r},t,s_z) = \psi(\mathbf{r}_1, \mathbf{r}_2,\ldots, \mathbf{r}_N,t)\xi(s_{z1}, s_{z2},\ldots,s_{zN}) = \phi(\mathbf{r}_1, \mathbf{r}_2,\ldots, \mathbf{r}_N)\zeta(s_{z1}, s_{z2},\ldots,s_{zN},t)\,.$
and for the case of identical particles, each factor has to have the correct antisymmetry or symmetry, to make the overall wave function antisymmetric for fermions or symmetric for bosons.
## Inner product
### Position-space inner products
The inner product of two wave functions Ψ1 and Ψ2 is useful and important for a number of reasons given below. For the case of one spinless particle in 1d, it can be defined as the complex number (at time t)[nb 9]
$\left\langle \Psi_1 , \Psi_2 \right\rangle = \int\limits_{-\infty}^\infty d x \, \Psi_1^*(x, t)\Psi_2(x, t).$
More generally, the formulae for the inner products are integrals over all coordinates or momenta and sums over all spin quantum numbers. That is, for one spinless particle in 3d the inner product of two wave functions can be defined as the complex number:
$\langle \Psi_1 , \Psi_2 \rangle = \int\limits_{\mathrm{ all \, space}} d^3\mathbf{r} \, \Psi_1^*(\mathbf{r},t)\Psi_2(\mathbf{r},t) \,,$
while for many spinless particles in 3d:
$\langle \Psi_1 , \Psi_2 \rangle = \int\limits_{\mathrm{ all \, space}} d ^3\mathbf{r}_1 \int\limits_{\mathrm{ all \, space}} d ^3\mathbf{r}_2\cdots \int\limits_{\mathrm{ all \, space}} d ^3\mathbf{r}_N \, \Psi_1^*(\mathbf{r}_1 \cdots \mathbf{r}_N,t)\Psi_2(\mathbf{r}_1 \cdots \mathbf{r}_N,t)$
(altogether, this is N three-dimensional volume integrals with differential volume elements d3ri, also written "dVi" or "dxi dyi dzi"). For one particle with spin in 3d:
$\langle \Psi_1 , \Psi_2 \rangle = \sum_{\mathrm{all\, }s_z} \int\limits_{\mathrm{ all \, space}} \, d^3\mathbf{r} \Psi^{*}_1(\mathbf{r},t,s_z)\Psi_2(\mathbf{r},t,s_z) \,,$
and for the general case of N particles with spin in 3d:
$\langle \Psi_1 , \Psi_2 \rangle = \sum_{s_{z\,N}} \cdots \sum_{s_{z\,2}} \sum_{s_{z\,1}} \int\limits_{\mathrm{ all \, space}} d ^3\mathbf{r}_1 \int\limits_{\mathrm{ all \, space}} d ^3\mathbf{r}_2\cdots \int\limits_{\mathrm{ all \, space}} d ^3 \mathbf{r}_N \Psi^{*}_1 \left(\mathbf{r}_1 \cdots \mathbf{r}_N,s_{z\,1}\cdots s_{z\,N},t \right )\Psi_2 \left(\mathbf{r}_1 \cdots \mathbf{r}_N,s_{z\,1}\cdots s_{z\,N},t \right )$
(altogether, N three-dimensional volume integrals followed by N sums over the spins).
In the Copenhagen interpretation, the modulus squared of the inner product (a complex number) gives a real number
$\left|\left\langle\Psi_1,\Psi_2\right\rangle\right|^2 = P\left(\Psi_2 \rightarrow \Psi_1\right) \,,$
which is interpreted as the probability of the wave function Ψ2 "collapsing" to the new wave function Ψ1 upon measurement of an observable, whose eigenvalues are the possible results of the measurement, with Ψ1 being an eigenvector of the resulting eigenvalue.
Although the inner product of two wave functions is a complex number, the inner product of a wave function Ψ with itself,
$\langle\Psi,\Psi\rangle = \|\Psi\|^2 \,,$
is always a positive real number. The number ||Ψ|| (not ||Ψ||2) is called the norm of the wave function Ψ, and is not the same as the modulus |Ψ|.
A wave function is normalized if:
$\left\langle \Psi, \Psi \right\rangle = 1 \,.$
If Ψ is not normalized, then dividing by its norm gives the normalized function Ψ/||Ψ||.
Two wave functions Ψ1 and Ψ2 are orthogonal if their inner product is zero:
$\left\langle \Psi_1, \Psi_2 \right\rangle = 0 \,.$
A set of wave functions Ψ1, Ψ2, ... are orthonormal if they are each normalized and are all orthogonal to each other:
$\langle \Psi_m , \Psi_n \rangle = \delta_{mn} \,,$
where m and n each take values 1, 2, ..., and δmn is the Kronecker delta (+1 for m = n and 0 for mn). Orthonormality of wave functions is instructive to consider since this guarantees linear independence of the functions. (However, the wave functions do not have to be orthonormal and can still be linearly independent, but the inner product of Ψm and Ψn is more complicated than the mere δmn).
Returning to the superposition above:
$\Psi = \sum_n a_n \psi_n$
if the basis wave functions ψn are orthonormal, then the coefficients have a particularly simple form:
$a_n = \langle \psi_n , \Psi \rangle$
If the basis wave functions were not orthonormal, then the coefficients would be different.
### Momentum-space inner products
Analogous to the position case, the inner product of two wave functions Φ1(p, t) and Φ2(p, t) can be defined as:
$\langle \Phi_1 , \Phi_2 \rangle = \int\limits_{-\infty}^\infty d p \, \Phi_1^*(p, t)\Phi_2(p, t) \,,$
and similarly for more particles in higher dimensions.
One particular solution to the time-independent Schrödinger equation is
$\Psi_p(x) = e^{ipx/\hbar},$
a plane wave, which can be used in the description of a particle with momentum exactly p, since it is an eigenfunction of the momentum operator. These functions are not normalizable to unity (they aren't square-integrable), so they are not really elements of physical Hilbert space. The set
$\{\Psi_p(x, t), -\infty \le p \le \infty\}$
forms what is called the momentum basis. This "basis" is not a basis in the usual mathematical sense. For one thing, since the functions aren't normalizable, they are instead normalized to a delta function,
$\langle\Psi_{p},\Psi_{p'}\rangle = \delta(p - p').$
For another thing, though they are linearly independent, there are too many of them (they form an uncountable set) for a basis for physical Hilbert space. They can still be used to express all functions in it using Fourier transforms as described above.
## Units of the wave function
Although wave functions are complex numbers, both the real and imaginary parts each have the same units (the imaginary unit i is a pure number without physical units). The units of ψ depend on the number of particles N the wave function describes, and the number of spatial or momentum dimensions n of the system.
When integrating |ψ|2 over all the coordinates, the volume element dnr1dnr2...dnrN has units of [length]Nn. Since the normalization conditions require the integral to be the unitless number 1, |ψ|2 must have units of [length]Nn, thus the units of |ψ| and hence ψ are [length]Nn/2. Likewise, in momentum space, length is replaced by momentum, and the units are [momentum]Nn/2. These results are true for particles of any spin, since for particles with spin, the summations are over dimensionless spin quantum numbers.
## More on wave functions and abstract state space
Main article: Quantum state
As has been demonstrated, the set of all possible normalizable wave functions for a system with a particular choice of basis constitute a Hilbert space. This vector space is in general infinite-dimensional. Due to the multiple possible choices of basis, these Hilbert spaces are not unique. One therefore talks about an abstract Hilbert space, state space, where the choice of basis is left undetermined. The choice of basis corresponds to a choice of a maximal set of quantum numbers, each quantum number corresponding to an observable. Two observables corresponding to quantum numbers in the maximal set must commute, therefore, the basis isn't entirely arbitrary, but nonetheless, there are always several choices.
Specifically, each state is represented as an abstract vector in state space[32]
$|\Psi\rangle$
where is a "ket" (a vector) written in Dirac's bra–ket notation.[33] Kets that differ by multiplication by a scalar represent the same state. A ray in Hilbert space is a set of normalized vectors differing by a complex number of modulus 1. If |ψ and |ϕ are two states in the vector space, and a and b are two complex numbers, then the linear combination
$|\Psi \rangle = a|\psi\rangle + b|\phi\rangle$
is also in the same vector space. The state space is postulated to have an inner product, denoted by
$\langle \Psi_1 | \Psi_2 \rangle,$
that is (usually, this differs) linear in the first argument and antilinear in the second argument. The dual vectors are denoted as "bras", Ψ|. These are linear functionals, elements of the dual space to the state space. The inner product, one chosen, can be used to define a unique map from state space to its dual, see Riesz representation theorem. this map is antilinear. One has
$\langle \Psi | = a^{*} \langle \psi | + b^{*} \langle \phi | \leftrightarrow a|\psi\rangle + b|\phi\rangle = |\Psi\rangle,$
where the asterisk denotes the complex conjugate. For this reason one has under this map
$\langle \Phi|\Psi\rangle = \langle \Phi| (|\Psi\rangle),$
and one may, as a practical consequence, at least notation-wise in this formalism, ignore that bra's are dual vectors.
The state vector for the system evolves in time according to the Schrödinger equation, or other dynamical pictures of quantum mechanics- In bra-ket notation this reads,
$i\hbar\frac{d}{dt}|\Psi\rangle = \hat{H} |\Psi\rangle$
Abstract state space is also, by definition, required to be a Hilbert space. The only requirement missing for this in the description so far is completeness. See the quantum state article for more explanation of the Hilbert space formalism and its consequences to quantum physics.
The connection to the Hilbert spaces of wave functions is made as follows. If (a, b, … l, m, …) is a maximal set of quantum numbers, denote the state corresponding to fixed choices of these quantum numbers by
$|a, b, \ldots, l, m, \ldots\rangle.$
The wave function corresponding to an arbitrary state is denoted
$\langle a, b, \ldots, l, m, \ldots|\Psi\rangle,$
for a concrete example,
$\Psi(x) = \langle x|\Psi\rangle.$
There are several advantages to understanding wave functions as representing elements of an abstract vector space:
• All the powerful tools of linear algebra can be used to manipulate and understand wave functions. For example:
• Linear algebra explains how a vector space can be given a basis, and then any vector in the vector space can be expressed in this basis. This explains the relationship between a wave function in position space and a wave function in momentum space, and suggests that there are other possibilities too.
• Bra–ket notation can be used to manipulate wave functions.
• The idea that quantum states are vectors in an abstract vector space (technically, a complex projective Hilbert space) is completely general in all aspects of quantum mechanics and quantum field theory, whereas the idea that quantum states are complex-valued "wave" functions of space is only true in certain situations.
Following is a summary of the bra–ket formalism applied to wave functions, with general discrete or continuous bases.
### Discrete and continuous bases
Discrete components Ak of a complex vector |A = ∑k Ak|ek, which belongs to a countably infinite-dimensional Hilbert space; there are countably infinitely many k values and basis vectors |ek.
Continuous components ψ(x) of a complex vector |ψ = ∫dx ψ(x)|x, which belongs to an uncountably infinite-dimensional Hilbert space; there are uncountably infinitely many x values and basis vectors |x.
Components of complex vectors plotted against index number; discrete k and continuous x. Two probability amplitudes out of infinitely many are highlighted.
A Hilbert space with a discrete basis |εi for i = 1, 2...n is orthonormal if the inner product of all pairs of basis kets are given by the Kronecker delta:
$\langle \varepsilon_i | \varepsilon_j \rangle = \delta_{ij}\,.$
Orthonormal bases are convenient to work with because the inner product of two vectors have simple expressions. A wave function expressed in this discrete basis of the Hilbert space, and the corresponding bra in the dual space, are respectively given by:
$| \Psi \rangle = \sum_{i = 1}^n c_i | \varepsilon_i \rangle = \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix} \,,\quad \langle \Psi | = | \Psi \rangle^{\dagger} = \sum_{i = 1}^n c^{*}_i \langle \varepsilon_i | = \begin{bmatrix} c_1^{*} & \cdots & c_n^{*} \end{bmatrix} \,,$
where the complex numbers
$c_i = \langle \varepsilon_i | \Psi \rangle$
are the components of the vector. The column vector is a useful way to list the numbers, and operations on the entire vector can be done according to matrix addition and multiplication. The entire vector is independent of the basis, but the components depend on the basis. If a change of basis is made, the components of the vector must also change to compensate.
A Hilbert space with a continuous basis { |ε } is orthonormal if the inner product of all pairs of basis kets are given by the Dirac delta function:
$\langle \varepsilon | \varepsilon' \rangle = \delta (\varepsilon-\varepsilon')\,.$
As with the discrete bases, a symbol ε is used in the basis states, two common notations are |ε and sometimes ε. A particular basis ket may be subscripted |ε0ε0 or primed |εε, or simply given another symbol in place of ε.
While discrete basis vectors are summed over a discrete index, continuous basis vectors are integrated over a continuous index (a variable of a function). In what follows, all integrals are with respect to the real-valued basis variable ε (not complex-valued), over the required range. Usually this is just the real line or subsets of it. The state in the continuous basis of the Hilbert space, with the corresponding bra in the dual space, are respectively given by:[34]
$| \Psi \rangle = \int d \varepsilon | \varepsilon \rangle \Psi(\varepsilon) \,,\quad \langle \Psi | = \int d \varepsilon \langle \varepsilon | {\Psi(\varepsilon)}^{*} \,,$
where the components are the complex-valued functions
$\Psi(\varepsilon) = \langle \varepsilon | \Psi \rangle$
of a real variable ε.
### Completeness conditions
The completeness conditions (also called closure relations) are
$\sum_{i=1}^n | \varepsilon_i \rangle \langle \varepsilon_i | = 1 \,,\quad \int d\varepsilon \, | \varepsilon \rangle \langle \varepsilon | = 1 \,$
for discrete and continuous orthonormal bases, respectively. An orthonormal set of kets form bases if and only if they satisfy these relations.[34] In each case, the equality to unity means this is an identity operator; its action on any state leaves it unchanged. Multiplying any state on the right of these gives the representation of the state in the basis. The inner product of a first state 1 with a second 2 can also be obtained by multiplying 1 on the left and 2 on the right of the relevant completeness condition.
### Inner product
Physically, the nature of the inner product is dependent on the basis in use, because the basis is chosen to reflect the quantum state of the system.
If 1 is a state in the above basis with components c1, c2, ..., cn and 2 is another state in the same basis with components z1, z2, ..., zn, the inner product is the complex number:
$\langle \Psi_1 | \Psi_2 \rangle = \left(\sum_i z^*_i \langle\varepsilon_i | \right)\left(\sum_j c_j |\varepsilon_j\rangle \right) =\sum_{ij} z^*_i c_j \langle \varepsilon_i|\varepsilon_j\rangle = \sum_i z^*_i c_i \,.$
If 1 is a state in the above continuous basis with components Ψ1(ε′), and 2 is another state in the same basis with components Ψ2(ε), the inner product is the complex number:
$\langle \Psi_1 | \Psi_2 \rangle = \left( \int d\varepsilon' {\Psi_1(\varepsilon')}^{*} \langle \varepsilon' | \right)\left(\int d\varepsilon \Psi_2(\varepsilon) |\varepsilon \rangle \right) = \int d\varepsilon' \int d\varepsilon {\Psi_1(\varepsilon')}^{*} \Psi_2(\varepsilon) \langle\varepsilon' | \varepsilon \rangle = \int d\varepsilon {\Psi_1(\varepsilon)}^{*} \Psi_2(\varepsilon) \,.$
where the integrals are taken over all ε and ε.
The square of the norm (magnitude) of the state vector is given by the inner product of with itself, a real number:
$\|\Psi\|^2 = \langle \Psi | \Psi \rangle = \sum_{j=1}^n | c_j |^2 \,,\quad \|\Psi\|^2 = \langle \Psi | \Psi \rangle = \int d \varepsilon \, | \Psi(\varepsilon) |^2$
for the discrete and continuous bases, respectively. Each say the projection of a complex probability amplitude onto itself is real. If is normalized, these expressions would be each separately equal to 1. If the state is not normalized, then dividing by its magnitude normalizes the state:
$| \Psi_N \rangle = \frac{1}{\| \Psi \|} | \Psi \rangle$
### Normalized components and probabilities
In the literature, the following results are often presented with normalized wavefunctions. Here, we keep the normalization factors to show where they appear if the wavefunction is not already normalized.
For the discrete basis, projecting the normalized state N onto a particular state the system may collapse to, |εq, gives the complex number;
$\langle \varepsilon_q | \Psi_N \rangle = \langle \varepsilon_q | \frac{1}{\|\Psi\|} \left ( \sum_{i = 1}^n c_i | \varepsilon_i \rangle \right ) = \frac{c_q}{\|\Psi\|} \,,$
so the modulus squared of this gives a real number;
$P(\varepsilon_q) = \left| \langle \varepsilon_q | \Psi_N \rangle \right|^2 = \frac{\left|c_q\right|^2}{\|\Psi\|^2} \, ,$
In the Copenhagen interpretation, this is the probability of state |εq occurring.
In the continuous basis, the projection of the normalized state onto some particular basis |ε is a complex-valued function;
$\langle \varepsilon' | \Psi_N \rangle = \langle \varepsilon' | \left( \frac{1}{\|\Psi\|}\int d \varepsilon | \varepsilon \rangle \Psi(\varepsilon) \right) = \frac{1}{\|\Psi\|}\int d \varepsilon \langle \varepsilon' | \varepsilon \rangle \Psi(\varepsilon) = \frac{1}{\|\Psi\|}\int d \varepsilon \delta( \varepsilon' - \varepsilon ) \Psi(\varepsilon) = \frac{\Psi(\varepsilon')}{\|\Psi\|} \,,$
so the squared modulus is a real-valued function
$\rho(\varepsilon') = \left| \langle \varepsilon' | \Psi_N \rangle \right|^2 = \frac{\left|\Psi(\varepsilon')\right|^2}{\|\Psi\|^2}$
In the Copenhagen interpretation, this function is the probability density function of measuring the observable ε, so integrating this with respect to ε between aε′ ≤ b gives:
$P_{a \leq \varepsilon \leq b} = \frac{1}{\|\Psi\|^2}\int_a^b d\varepsilon' | \Psi(\varepsilon') |^2 = \frac{1}{\|\Psi\|^2}\int_a^b d\varepsilon' | \langle \varepsilon' | \Psi \rangle |^2 \,,$
the probability of finding the system with ε between ε′ = a and ε′ = b.
### Wave function collapse
The physical meaning of the components of is given by the wave function collapse postulate, also known as wave function collapse. If the observable(s) ε (momentum and/or spin, position and/or spin, etc.) corresponding to states |εi has distinct and definite values, λi, and a measurement of that variable is performed on a system in the state then the probability of measuring λi is |εi|2. If the measurement yields λi, the system "collapses" to the state |εi irreversibly and instantaneously.
### Time dependence
In the Schrödinger picture, the states evolve in time, so the time dependence is placed in according to[35]
$|\Psi(t)\rangle = \sum_i \, | \varepsilon_i \rangle \langle \varepsilon_i | \Psi(t)\rangle = \sum_i c_i(t) | \varepsilon_i \rangle$
for discrete bases, or
$|\Psi(t)\rangle = \int d\varepsilon \, | \varepsilon \rangle \langle \varepsilon | \Psi(t)\rangle = \int d\varepsilon \, \Psi(\varepsilon,t) | \varepsilon \rangle$
for continuous bases. However, in the Heisenberg picture the states are constant in time and time dependence is placed in the Heisenberg operators, so is not written as |Ψ(t). The Heisenberg picture wave function is a snapshot of a Schrödinger picture wave function, representing the whole spacetime history of the system. In the interaction picture (also called Dirac picture), the time dependence is placed in both the states and operators, the subdivision depending on the interaction term in the Hamiltonian, and can be viewed as intermediate between the Heisenberg and Schrödinger pictures. It is useful primarily in computing S-matrix elements.[36]
### Tensor product
It is useful to introduce another operation with the physical interpretation of forming composite states from a collection of other states. This is the tensor product. Given two systems described by states and , the tensor product of the states forms the composite state denoted by or simply without any operation symbol , and the new system includes both of the original systems together. The tensor product state lives in a new space; the tensor product of the original Hilbert spaces. The bases spanning this space are the tensor products of the original bases. The product is not commutative in general, so . If has components ci and has components zj, each in a discrete orthonormal basis |εk, then:
$|\Psi\rangle|\Phi\rangle = \left(\sum_i c_i |\varepsilon_i\rangle \right) \left(\sum_j z_j |\varepsilon_j\rangle \right) = \sum_{i,j}c_iz_j|\varepsilon_i\rangle|\varepsilon_j\rangle$
and the notation can be simplified by abbreviating |A = , Aij = cizj, and |Eij = |εi|εj, so that
$|A\rangle = \sum_{i,j}A_{ij} |E_{ij}\rangle$
The same procedure follows for continuous bases using integration. This can also be extended to any number of states, however taking tensor products for fermions and bosons is complicated by the symmetry requirements, see identical particles for general results.
## Position representations
This section applies mostly to non-relativistic quantum mechanics. In relativistic quantum mechanics, eigenstates of the position operator are problematic due to a relativistic extension of Heisenberg's uncertainty principle. In relativistic quantum field theory, they are not used at all to label physical states. Associated to a particle perfectly localized to a point in space is an infinite uncertainty in energy. This leads to pair production in the relativistic regime. Thus such a particle automatically has companions, leading to a breakdown of the description.
### State space for one spin-0 particle in 1d
For a spinless particle in one spatial dimension (the x-axis or real line), the state can be expanded in terms of a continuum of basis states; |x, also written x, corresponding to the set of all position coordinates x. The completeness condition for this basis is
$1 = \int\limits _{-\infty}^{\infty} d x \, | x \rangle \langle x |$
and the orthogonality relation is
$\langle x' | x \rangle = \delta(x'-x)$
The state is expressed by:
$| \Psi \rangle = \left(\int\limits _{-\infty}^{\infty} d x \, | x \rangle \langle x |\right)| \Psi \rangle = \int\limits _{-\infty}^{\infty} d x \, | x \rangle \langle x | \Psi \rangle = \int\limits _{-\infty}^{\infty} d x \, \Psi (x) | x \rangle$
in which the "wave function" described as a function is a component of the complex state vector.
$\Psi (x) = \langle x | \Psi \rangle$
$\langle \Psi_1 | \Psi_2 \rangle = \langle \Psi_1 |\left(\int\limits_{-\infty}^{\infty} d x \, | x \rangle \langle x | \right)| \Psi_2 \rangle = \int\limits_{-\infty}^{\infty} d x \, \langle \Psi_1 | x \rangle \langle x | \Psi_2 \rangle = \int\limits _{-\infty}^{\infty} d x \, \Psi_1(x)^{*} \Psi_2(x) \,.$
If the particle is confined to a region R (a subset of the x-axis), the integrals in the inner product and completeness condition would be integrals over R.
### State space (other cases)
The previous example can be extended to more particles in higher dimensions, and include spin.
For one spinless particle in 3d, the basis states are |r and any state vector in this space is expressed in terms of the basis vectors as |r:
$|\Psi\rangle = \int\limits_\mathrm{all\,space} d^3\mathbf{r} |\mathbf{r}\rangle \langle \mathbf{r} | \Psi\rangle$
with components:
$\langle \mathbf{r} | \Psi\rangle = \Psi(\mathbf{r})$
For N spinless particles in 3d, the basis states are |r1, ..., rN. This is the tensor product of the one-particle position bases |r1, |r2, ..., |rN, each of which spans the separate one-particle Hilbert spaces, so |r1, ..., rN are the basis states for the tensor product of the one-particle Hilbert spaces (the Hilbert space for the composite many particle system). Any state vector in this space is
$|\Psi\rangle = \int\limits_\mathrm{all \, space} d^3\mathbf{r}_N\cdots \int\limits_\mathrm{all \, space} d^3\mathbf{r}_2 \int\limits_\mathrm{all \, space} d^3\mathbf{r}_1 | \mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N \rangle \langle \mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N | \Psi\rangle$
with components:
$\langle \mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N | \Psi \rangle = \Psi(\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N)$
For one particle with spin in 3d, the basis states are |r, sz, the tensor product of the position basis |r and spin basis |sz, which exists in a new space from the spin space and position space alone. Any state in this space is:
$|\Psi\rangle = \sum_{s_z} \int\limits_\mathrm{all\,space} d^3\mathbf{r} |\mathbf{r},s_z\rangle \langle \mathbf{r},s_z | \Psi\rangle$
with components:
$\langle \mathbf{r},s_z | \Psi\rangle = \Psi(\mathbf{r}, s_z )$
For N particles with spin in 3d, the basis states are |r1, ..., rN, sz 1, ..., sz N, the tensor product of the position basis |r1, ..., rN and spin basis |sz 1, ..., sz N, which exists in a new space from the spin space and position space alone. Any state in this space is:
$|\Psi\rangle = \sum_{s_{z\,1} , \ldots , s_{z\,N}}\int\limits\limits_\mathrm{all\,space} d^3\mathbf{r}_N \cdots \int\limits\limits_\mathrm{all\,space} d^3\mathbf{r}_1 \, | \mathbf{r}_1, \ldots, \mathbf{r}_N , s_{z\,1} , \ldots , s_{z\,N} \rangle \langle \mathbf{r}_1, \ldots, \mathbf{r}_N , s_{z\,1} , \ldots , s_{z\,N} | \Psi\rangle$
with components:
$\langle \mathbf{r}_1, \ldots, \mathbf{r}_N , s_{z\,1} , \ldots , s_{z\,N} | \Psi\rangle = \Psi(\mathbf{r}_1, \ldots, \mathbf{r}_N , s_{z\,1} , \ldots , s_{z\,N} )$
If the particles are restricted to regions of position space, then the integrals in the completeness relations are taken over those regions, rather than the entire coordinate space. For the general case of many particles with spin in 3d, if particle 1 is in region R1, particle 2 is in region R2, and so on, the state in this position–spin representation is:
$| \Psi \rangle = \sum_{s_{z\,1} , \ldots , s_{z\,N}}\int\limits\limits_{R_N} d^3\mathbf{r}_N \cdots \int\limits\limits_{R_1} d^3\mathbf{r}_1 \, \Psi ( \mathbf{r}_1, \ldots, \mathbf{r}_N , s_{z\,1} , \ldots , s_{z\,N} ) | \mathbf{r}_1, \ldots, \mathbf{r}_N , s_{z\,1} , \ldots , s_{z\,N} \rangle$
The orthogonality relation for this basis is:
$\langle \mathbf{x}_1, \ldots, \mathbf{x}_{N}, m_1, \ldots, m_N | \mathbf{r}_1,\ldots,\mathbf{r}_N, s_{z\,1},\ldots,s_{z\,N} \rangle = \delta_{m_1\,s_{z\,1}}\cdots\delta_{m_N\,s_{z\,N}}\delta( \mathbf{x}_1 - \mathbf{r}_{1})\cdots\delta( \mathbf{x}_{N} - \mathbf{r}_{N})$
and the inner product of 1 and 2 is:
$\langle \Psi_1 | \Psi_2 \rangle = \sum_{s_{z\,1} , \ldots , s_{z\,N}}\int\limits\limits_{R_N} d^3\mathbf{r}_N \cdots \int\limits\limits_{R_1} d^3\mathbf{r}_1 \Psi_1(\mathbf{r}_1, \ldots, \mathbf{r}_N , s_{z\,1} , \ldots , s_{z\,N} )^{*} \Psi_2(\mathbf{r}_1, \ldots, \mathbf{r}_N, s_{z\,1} , \ldots , s_{z\,N}) \,.$
Momentum space wave functions are similar, using the momentum vectors of the particles as continuous bases, namely |p, |p1, p2, ..., pN, etc.
## Ontology
Whether the wave function really exists, and what it represents, are major questions in the interpretation of quantum mechanics. Many famous physicists of a previous generation puzzled over this problem, such as Schrödinger, Einstein and Bohr. Some advocate formulations or variants of the Copenhagen interpretation (e.g. Bohr, Wigner and von Neumann) while others, such as Wheeler or Jaynes, take the more classical approach[37] and regard the wave function as representing information in the mind of the observer, i.e. a measure of our knowledge of reality. Some, including Schrödinger, Bohm and Everett and others, argued that the wave function must have an objective, physical existence. Einstein thought that a complete description of physical reality should refer directly to physical space and time, as distinct from the wave function, which refers to an abstract mathematical space.[38]
## Examples
### Free particle
Main article: Free particle
A free particle in 3d with wave vector k and angular frequency ω has a wave function
$\Psi (\mathbf{r},t) = A e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\,.$
### Particle in a box
Main article: Particle in a box
A particle is restricted to a 1D region between x = 0 and x = L; its wave function is:
\begin{align} \Psi (x,t) & = Ae^{i(kx-\omega t)}, & 0 \leq x \leq L \\ \Psi (x,t) & = 0, & x < 0, x > L \\ \end{align}
To normalize the wave function we need to find the value of the arbitrary constant A; solved from
$\int\limits_{-\infty}^{\infty} dx \, |\Psi|^2 = 1 .$
From Ψ, we have |Ψ|2 = A2, so the integral becomes;
$\int\limits_{-\infty}^0 dx \cdot 0 + \int\limits_0^L dx \, A^2 + \int\limits_L^\infty dx \cdot 0 = 1 ,$
Solving this equation gives A = 1/L, so the normalized wave function in the box is;
$\Psi (x,t) = \frac{1}{\sqrt{L}} e^{i(kx-\omega t)}, \quad 0 \leq x \leq L\,.$
### One-dimensional quantum tunnelling
Scattering at a finite potential barrier of height $V_0$ The amplitudes and direction of left and right moving waves are indicated. In red, those waves used for the derivation of the reflection and transmission amplitude. $E > V_0$ for this illustration.
One of most prominent features of the wave mechanics is a possibility for a particle to reach a location with a prohibitive (in classical mechanics) force potential. In the one-dimensional case of particles with energy less than $V_0$ in the square potential
$V(x)=\begin{cases}V_0 & |x|
the steady-state solutions to the wave equation have the form (for some constants $k, \kappa$)
$\psi (x) = \begin{cases} A_{\mathrm{r}}\exp(ikx)+A_{\mathrm{l}}\exp(-ikx) & x<-a, \\ B_{\mathrm{r}}\exp(\kappa x)+B_{\mathrm{l}}\exp(-\kappa x) & |x|\le a, \\ C_{\mathrm{r}}\exp(ikx)+C_{\mathrm{l}}\exp(-ikx) & x>a. \end{cases}$
Note that these wave functions are not normalized; see scattering theory for discussion.
The standard interpretation of this is as a stream of particles being fired at the step from the left (the direction of negative x): setting Ar = 1 corresponds to firing particles singly; the terms containing Ar and Cr signify motion to the right, while Al and Cl – to the left. Under this beam interpretation, put Cl = 0 since no particles are coming from the right. By applying the continuity of wave functions and their derivatives at the boundaries, it is hence possible to determine the constants above.
### Other
Some examples of wave functions for specific applications include:
## Remarks
1. ^ For this statement to make sense, the observables need to be elements of a maximal commuting set. To see this, it is a simple matter to note that, for example, the momentum operator of the i'th particle in an n-particle system is not a generator of any symmetry in nature. On the other hand, the total angular momentum is a generator of a symmetry in nature; the translational symmetry.
2. ^ The resulting basis may or may not technically be a basis in the mathematical sense of Hilbert spaces. For instance, states of definite position and definite momentum are not square integrable. This may be overcome with the use of wave packets or by enclosing the system in a "box". See further remarks below.
3. ^ In technical terms, this is formulated the following way. The inner product yields a norm. This norm in turn induces a metric. If this metric is complete, then the aforementioned limits will be in the function space. The inner product space is then called complete. A complete inner product space is a Hilbert space. The abstract state space is always taken as a Hilbert space. The matching requirement for the function spaces is a natural one. The Hilbert space property of the abstract state space was originally extracted from the observation that the function spaces forming normalizable solutions to the Schrödinger equation are Hilbert spaces.
4. ^ As is explained in a later footnote, the integral must be taken to be the Lebesgue integral, the Riemann integral is not sufficient.
5. ^ One such relaxation is that the wave function must belong to the Sobolev space W1,2. It means that it is differentiable in the sense of distributions, and its gradient is square-integrable. This relaxation is necessary for potentials that are not functions but are distributions, such as the Dirac delta function.
6. ^ It is easy to visualize a sequence of functions meeting the requirement that converges to a discontinuous function. For this, modify an example given in Inner product space#Examples. This element though is an element of L2.
7. ^ For instance, in perturbation theory one may construct a sequence of functions approximating the true wave function. This sequence will be guaranteed to converge in a larger space, but without the assumption of a full-fledged Hilbert space, it will not be guaranteed that the convergence is to a function in the relevant space and hence solving the original problem.
8. ^ Some functions not being square-integrable, like the plane-wave free particle solutions are necessary for the description as outlined in a previous note and also further below.
9. ^ The functions are here assumed to be elements of L2, the space of square integrable functions. The elements of this space are more precisely equivalence classes of square integrable functions, two functions declared equivalent if they differ on a set of Lebesgue measure 0. This is necessary to obtain an inner product (that is, (Ψ, Ψ) = 0 ⇒ Ψ ≡ 0) as opposed to a semi-inner product. The integral is taken to be the Lebesque integral. This is essential for completeness of the space, thus yielding a complete inner product space = Hilbert space.
## Notes
1. ^ Born 1927, pp. 354–357
2. ^ Heisenberg 1958, p. 143
3. ^ Heisenberg, W. (1927/1985/2009). Heisenberg is translated by Camilleri 2009, p. 71, (from Bohr 1985, p. 142).
4. ^ Murdoch 1987, p. 43
5. ^ de Broglie 1960, p. 48
6. ^ Landau Lifshitz, p. 6
7. ^ Newton 2002, pp. 19–21
8. ^ Lerner & Trigg 1991, pp. 1223–1229
9. ^ Einstein 1905, pp. 132–148 (in German), Arons & Peppard 1965, p. 367 (in English)
10. ^ Einstein 1916, pp. 47–62 and a nearly identical version Einstein 1917, pp. 121–128 translated in ter Haar 1967, pp. 167–183.
11. ^ de Broglie 1923, pp. 507–510,548,630
12. ^ Hanle 1977, pp. 606–609
13. ^ Schrödinger 1926, pp. 1049–1070
14. ^ Tipler, Mosca & Freeman 2008
15. ^ a b c Weinberg 2013
16. ^ a b c Born 1926a, translated in Wheeler & Zurek 1983 at pages 52–55.
17. ^ Born 1926b, translated in Ludwig 1968, pp. 206–225. Also here.
18. ^ Young & Freedman 2008, p. 1333
19. ^ a b c Atkins 1974
20. ^ Martin & Shaw 2008
21. ^ Pauli 1927, pp. 601–623.
22. ^ Weinberg (2002) takes the standpoint that quantum field theory appears the way it does because it is the only way to reconcile quantum mechanics with special relativity.
23. ^ Weinberg (2002) See especially chapter 5, where some of these results are derived.
24. ^ Weinberg 2002 Chapter 4.
25. ^ Weinberg 2002
26. ^ Eisberg & Resnick 1985
27. ^ Rae 2008
28. ^ Atkins 1974, p. 258
29. ^ a b Griffiths 2004
30. ^ Zettili 2009, p. 463
31. ^ Shankar 1994, p. 378–379
32. ^ Dirac 1982
33. ^ Dirac 1939
34. ^ a b (Peleg et al. 2010) pp. 64–65.
35. ^ (Peleg et al. 2010, pp. 68–69)
36. ^ Weinberg 2002 Chapter 3, Scattering matrix.
37. ^ Jaynes 2003
38. ^ Einstein 1998, p. 682
## References
• Bohr, N. (1985). J. Kalckar, ed. Niels Bohr - Collected Works: Foundations of Quantum Physics I (1926 - 1932) 6. Amsterdam: North Holland. ISBN 9780444532893.
• Born, M. (1926a). "Zur Quantenmechanik der Stossvorgange". Z. f. Physik 37: 863–867. doi:10.1007/bf01397477.
• Born, M. (1926b). "Quantenmechanik der Stossvorgange". Z. f. Physik 38: 803–827. doi:10.1007/bf01397184.
• de Broglie, L. (1960). Non-linear Wave Mechanics: a Causal Interpretation. Amsterdam: Elsevier.
• Camilleri, K. (2009). Heisenberg and the Interpretation of Quantum Mechanics: the Physicist as Philosopher. Cambridge UK: Cambridge University Press. ISBN 978-0-521-88484-6.
• Dirac, P. A. M. (1982). The principles of quantum mechanics. The international series on monographs on physics (4th ed.). Oxford University Press. ISBN 0 19 852011 5.
• Einstein, A. (1916). "Zur Quantentheorie der Strahlung". Mitteilungen der Physikalischen Gesellschaft Zürich 18: 47–62.
• Einstein, A. (1998). P. A. Schlipp, ed. Albert Einstein: Philosopher-Scientist. The Library of Living Philosophers VII (3rd ed.). La Salle Publishing Company, Illinois: Open Court. ISBN 0-87548-133-7.
• Eisberg, R.; Resnick, R. (1985). Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles (2nd ed.). John Wiley & Sons. ISBN 978-0-471-87373-0.
• Griffiths, D. J. (2004). Introduction to Quantum Mechanics (2nd ed.). Essex England: Pearson Education Ltd. ISBN 978-0131118928.
• Heisenberg, W. (1958). Physics and Philosophy: the Revolution in Modern Science. New York: Harper & Row.
• Hanle, P.A. (1977), "Erwin Schrodinger's Reaction to Louis de Broglie's Thesis on the Quantum Theory.", Isis 68 (4), doi:10.1086/351880
• Lerner, R.G.; Trigg, G.L. (1991). Encyclopaedia of Physics (2nd ed.). VHC Publishers. ISBN 0-89573-752-3.
• Murdoch, D. (1987). Niels Bohr's Philosophy of Physics. Cambridge UK: Cambridge University Press. ISBN 0-521-33320-2.
• Newton, R.G. (2002). Quantum Physics: a Text for Graduate Student. New York: Springer. ISBN 0-387-95473-2.
• Pauli, Wolfgang (1927). "Zur Quantenmechanik des magnetischen Elektrons". Zeitschrift für Physik (in German) 43.
• Peleg, Y.; Pnini, R.; Zaarur, E.; Hecht, E. (2010). Quantum mechanics. Schaum's outlines (2nd ed.). McGraw Hill. ISBN 978-0-07-162358-2.
• Shankar, R. (1994). Principles of Quantum Mechanics (2nd ed.). ISBN 0306447908.
• Martin, B.R.; Shaw, G. (2008). Particle Physics. Manchester Physics Series (3rd ed.). John Wiley & Sons. ISBN 978-0-470-03294-7.
• Tipler, P. A.; Mosca, G.; Freeman (2008). Physics for Scientists and Engineers – with Modern Physics (6th ed.). ISBN 0-7167-8964-7.
• Weinberg, S. (2002), The Quantum Theory of Fields 1, Cambridge University Press, ISBN 0-521-55001-7
• Young, H. D.; Freedman, R. A. (2008). Pearson, ed. Sears' and Zemansky's University Physics (12th ed.). Addison-Wesley. ISBN 978-0-321-50130-1. | 19,118 | 74,195 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 98, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2015-14 | latest | en | 0.904743 |
https://de.mathworks.com/matlabcentral/cody/problems/44785-lunar-arithmetic-addition/solutions/2613520 | 1,596,912,885,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738015.38/warc/CC-MAIN-20200808165417-20200808195417-00245.warc.gz | 269,020,586 | 15,864 | Cody
# Problem 44785. Lunar Arithmetic (Addition)
Solution 2613520
Submitted on 25 Jun 2020
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
x = 5; y = 6; assert(isequal(lunarAddition(x,y),6))
Conversion to double from cell is not possible. Error in lunarAddition (line 4) a(i) = varargin(i); Error in Test1 (line 3) assert(isequal(lunarAddition(x,y),6))
2 Fail
x = 456; y = 789; assert(isequal(lunarAddition(x,y),789))
Conversion to double from cell is not possible. Error in lunarAddition (line 4) a(i) = varargin(i); Error in Test2 (line 3) assert(isequal(lunarAddition(x,y),789))
3 Fail
x = 86; y = 12374; assert(isequal(lunarAddition(x,y),12386))
Conversion to double from cell is not possible. Error in lunarAddition (line 4) a(i) = varargin(i); Error in Test3 (line 3) assert(isequal(lunarAddition(x,y),12386))
4 Fail
x = 29; y = 1652; z = 95412; assert(isequal(lunarAddition(x,y,z),95659))
Conversion to double from cell is not possible. Error in lunarAddition (line 4) a(i) = varargin(i); Error in Test4 (line 4) assert(isequal(lunarAddition(x,y,z),95659))
5 Fail
x = 33; y = 1111; z = 4456; a = 38; assert(isequal(lunarAddition(x,y,z,a),4458))
Conversion to double from cell is not possible. Error in lunarAddition (line 4) a(i) = varargin(i); Error in Test5 (line 5) assert(isequal(lunarAddition(x,y,z,a),4458))
6 Fail
x = 85214; y = 4785; z = 1; a = 850615; b = 14702140; assert(isequal(lunarAddition(x,y,z,a,b),14885785))
Conversion to double from cell is not possible. Error in lunarAddition (line 4) a(i) = varargin(i); Error in Test6 (line 6) assert(isequal(lunarAddition(x,y,z,a,b),14885785))
7 Fail
x = 9; y = 0; assert(isequal(lunarAddition(x,y),9))
Conversion to double from cell is not possible. Error in lunarAddition (line 4) a(i) = varargin(i); Error in Test7 (line 3) assert(isequal(lunarAddition(x,y),9)) | 652 | 1,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-34 | latest | en | 0.638959 |
https://gymnasium.farama.org/environments/toy_text/cliff_walking/ | 1,716,749,332,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00052.warc.gz | 237,702,969 | 11,785 | # Cliff Walking#
This environment is part of the Toy Text environments which contains general information about the environment.
Action Space
`Discrete(4)`
Observation Space
`Discrete(48)`
import
`gymnasium.make("CliffWalking-v0")`
Cliff walking involves crossing a gridworld from start to goal while avoiding falling off a cliff.
## Description#
The game starts with the player at location [3, 0] of the 4x12 grid world with the goal located at [3, 11]. If the player reaches the goal the episode ends.
A cliff runs along [3, 1..10]. If the player moves to a cliff location it returns to the start location.
The player makes moves until they reach the goal.
Adapted from Example 6.6 (page 132) from Reinforcement Learning: An Introduction by Sutton and Barto [1].
## Action Space#
The action shape is `(1,)` in the range `{0, 3}` indicating which direction to move the player.
• 0: Move up
• 1: Move right
• 2: Move down
• 3: Move left
## Observation Space#
There are 3 x 12 + 1 possible states. The player cannot be at the cliff, nor at the goal as the latter results in the end of the episode. What remains are all the positions of the first 3 rows plus the bottom-left cell.
The observation is a value representing the player’s current position as current_row * nrows + current_col (where both the row and col start at 0).
For example, the stating position can be calculated as follows: 3 * 12 + 0 = 36.
The observation is returned as an `int()`.
## Starting State#
The episode starts with the player in state `[36]` (location [3, 0]).
## Reward#
Each time step incurs -1 reward, unless the player stepped into the cliff, which incurs -100 reward.
## Episode End#
The episode terminates when the player enters state `[47]` (location [3, 11]).
## Information#
`step()` and `reset()` return a dict with the following keys:
• “p” - transition proability for the state.
As cliff walking is not stochastic, the transition probability returned always 1.0.
## Arguments#
```import gymnasium as gym
gym.make('CliffWalking-v0')
```
## References#
[1] R. Sutton and A. Barto, “Reinforcement Learning: An Introduction” 2020. [Online]. Available: http://www.incompleteideas.net/book/RLbook2020.pdf
## Version History#
• v0: Initial version release | 576 | 2,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.878846 |
https://math.stackexchange.com/questions/3157905/prove-if-c28-equiv-0-mod-p-then-c3-7c2-8c-is-a-quadratic-residue-mod | 1,638,144,020,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00229.warc.gz | 474,371,544 | 33,607 | # Prove: if $c^2+8 \equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
I want to show:
If $$c^2+8 \equiv 0$$ mod $$p$$ for prime $$p>3$$, then $$c^3-7c^2-8c$$ is a quadratic residue mod $$p$$.
I have calculated that $$c^3-7c^2-8c \equiv -7c^2-16c \equiv 56- 16c \equiv 8(7-2c) \equiv c^2 (2c -7)$$, so it should be enough to see that $$2c -7$$ is a quadratic residue. What now?
$$2c-7\equiv2c-7+c^2+8\pmod p\equiv(c+1)^2$$
• @AJ. Have you noticed my other answer. That is much more natural derivation Mar 22 '19 at 9:10
We have that $$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) \equiv 2c - 7 \mod p.$$ This proves the claim.
Hint:
As $$c^2\equiv-8\pmod p,$$
$$c(c+1)=c^2+c\equiv c-8\pmod p$$
$$c(c+1)(c-8)\equiv?$$
I believe this is how the problem naturally came into being .
• Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different. Mar 22 '19 at 9:21
• Me neither... $(+1)$ Mar 24 '19 at 6:12 | 424 | 1,036 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-49 | latest | en | 0.861832 |
https://www.enotes.com/homework-help/which-one-following-an-example-balanced-chemical-732741 | 1,624,472,463,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00250.warc.gz | 686,861,910 | 18,438 | Which one of the following is an example of a balanced chemical reaction? A) `HCl + KMnO_4 = Cl_2 + MnO_2 + H_2O + KCl` B) `HCl + KMnO_4 = Cl_2 + MnO_2 + 2H_2O + KCl` C) `2HCl + 2KMnO_4 = Cl_2 + MnO_2 + 2H_2O + 2KCl` D) `6HCl + 2KMnO_4 = 2Cl_2 + 2MnO_2 + 4H_2O + 2KCl` E) `8HCl + 2KMnO_4 = 3Cl_2 + 2MnO_2 + 4H_2O + 2KCl`
Hello!
This is actually a math question.
For a chemical equation to be balanced, there must be an equal quantity of atoms of each type on both sides of an equation. Let's start from the first type of atoms involved here, `H` (hydrogen). It appears only in `H Cl` at the left (one atom for each molecule) and in `H_2O` at the right (two atoms for each molecule). Check:
A) `1*1 = 1*2` (false)
B) `1*1 = 2*2` (false)
C) `2*1 = 2*2` (false)
D) `6*1 = 4*2` (false)
E) `8*1 = 4*2` (true).
So the only equation which might be balanced is E. Check the remaining types of atoms: `Cl,` `K,` `Mn,` `O.`
`Cl:` `8*1 = 3*2 + 2*1` (true),
`K:` `2*1 = 2*1` (true),
`Mn:` `2*1 = 2*1` (true),
`O:` `2*4 = 2*2 + 4*1` (true).
Great, E is a balanced equation and the only such. The answer is E.
Approved by eNotes Editorial Team | 501 | 1,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-25 | latest | en | 0.687769 |
https://followinglearning.blogspot.com/2018/11/ | 1,723,404,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00298.warc.gz | 203,757,736 | 25,910 | ## Tuesday 27 November 2018
### Hexagons
Rather than triangles, which was on the Grade 3 (English Year 4) plan, I felt that hexagons would give us more scope for focusing on properties. It's worth looking at Christopher Danielson's lessons on this. I wasn't really expecting to get on to proof, but if a conjecture came up, I would be happy (it didn't this time!). We started with a Which One Doesn't Belong?
I mentioned that three of them were hexagons; they didn't have to be regular to be hexagons.
We asked the classes to make hexagons from pattern blocks, to make one whose shape was not chosen by anyone else. This had advantages. The hexagon, kind of, 'belonged' to the student. We got a range of different hexagons that we could construct and measure easily, and would also, mostly, tile. But I'm still wondering about the gains and losses from this approach, and I'll come back to that.
Finally, after a number of repeats and non-hexagons that we didn't spot straight away, we had a set:
which I reproduced for the class without the colour, so that we'd be focusing on the component shapes less:
- still time for another pattern block hexagon WODB though:
Notice that 'looks most like a hexagon'! We were going to shake off that regular hexagon thing...
We looked at side length, counting the length of almost every side of the pattern blocks as one. I also talked about perimeter. (BTW: my opinion is that it's really good to introduce this separately to area. Although it goes together with area in our minds like salt-and-pepper, my feeling is that some students get a little which-is-which?) We documented this for our hexagon, later adding on parallel lines and symmetry:
Then, starting from the known fact that a square's corner is 90°, I asked the class to find the angles on the other pattern blocks, which, together, step-by-step, they did. The big breakthough was when a pair realised that three of the acute angles of the thin rhombus fit into the squares corner. Then that was a measure for all the others.
So, the next day, we applied this to our hexagons. I'm pleased about this part of the sequence, because it seemed to make it possible for everyone.
Time for another WODB then next day, with four of our hexagons with angles shown.
I hesitate to add this bit, because I'm unusually lucky to have a colleague in secondary who laser cuts me shapes out of acrylic, and that's not accessible to many. But the kids were really excited when these arrived...
Immediately they began tessellating them:
We recorded this later:
This one AK couldn't get to tile:There's a couple we had difficulty tiling:
Martin helped out with one:
And Hana helped out with the other with a tiling that includes some other pattern block shapes:
Rod did the same:
All this led on nicely to our other tiling work with the pattern blocks.
To review what we'd done, I made little books about the 25 hexagons (taking too long on all the images, which was useful in class but I don't recommend to others!) and asked children to write about different aspects of them.
The books showed that the students had a lot to say about the various properties we'd looked at.
We also finished off with a WODB that students annotated individually. This told a mixed story. I'd chosen some hexagons that weren't within the pattern block family. The students did notice lots, and showed their knowledge through their annotations:
But some of them also treated these hexagons as if they had the same uniformity that the pattern block hexagons had. Some saw, for instance, the diagonal lines of the green shape as parallel. Some tried to see the hexagons as composed of pattern block shapes. And some tried to measure the perimeters - oh dear, I hadn't thought of that, and hadn't built in whole-number lengths! I kind of feel this was unfair of me, but it has made me more aware of something.
It seems we have some things to unlearn now. Another thing: for our brief estimation actiivity the following week I used Estimation 180's What degrees of the pie have been eaten? So many students put 90° or 60° - and suddenly I realised, now they think all angles are going to be in mutliples of 30°!
So we may need to unlearn:
• all lines that look vaguely parallel are parallel;
• all hexagons are made up of pattern block shapes;
• all side lengths are whole numbers;
• all angles are multiples of 30°.
Luckily, it's not hard to find counter-examples that undo most of these overgeneralisatons that I've led them into. The undoing could even be an impactful way of learning these important points. So, I'm not too disheartened.
But I'm still mulling this over. Would it have been better to have given a selection of more varied, non-pattern-block hexagons, like Christopher Danielson's
source
They would have lost the personal creation, the ease of measuring, the tiling perhaps. But they would have had a bigger, more representative example space.
Or I could have chosen a different playground for hexagons. A geoboard maybe, or as in Lana Pavlova's lovely hexagon work, the tangram. Or moved more quickly through differrent embodiments of hexagon.
All this connects with some of my recent thinking about examples and generalisation. Have a look at this, from Thinkers by Chris Bills, Liz Bills, John Mason and Anne Watson:
I'm now much more aware of how our choice of example, the variation we include, carries its own potential misinformation. Students at the start of this felt that a hexagon was a regular hexagon - that's what they'd been shown. We show a range of rectangles and don't include a square. We call the blue and light brown pattern blocks rhombuses, but (natually) call the square a square, even though it's a rhombus too. We look at fractions as circles. Students are going to see features of the examples we present as essential to the properties we want them to become familiar with. Some of those features will be, but others won't.
So the 'problem' with my hexagon lesson is perhaps a more general one...
How would you do a set of lessons on hexagons? What do you think about this problem with examples? As usual, I'd value any comments. | 1,401 | 6,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.9827 |
https://studentsblog100.blogspot.com/2015/06/ce6505-design-of-reinforced-concrete-elements-syllabus.html | 1,566,284,833,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315258.34/warc/CC-MAIN-20190820070415-20190820092415-00251.warc.gz | 654,069,703 | 37,532 | CE6505 DESIGN OF REINFORCED CONCRETE ELEMENTS SYLLABUS FOR 5TH SEM CIVIL REG 2013 - Anna University Internal marks 2018
ANNA UNIVERSITY CIVIL SYLLABUS
CE6505 DESIGN OF REINFORCED CONCRETE ELEMENTS SYLLABUS
5TH SEMESTER CIVIL ENGINEERING
REGULATION 2013
OBJECTIVES:
• To introduce the different types of philosophies related to design of basic structural elements such as slab, beam, column and footing which form part of any structural system with reference to Indian standard code of practice.
UNIT I METHODS OF DESIGN OF CONCRETE STRUCTURES
Concept of Elastic method, ultimate load method and limit state method – Advantages of Limit State Method over other methods – Design codes and specification – Limit State philosophy as detailed in IS code – Design of beams and slabs by working stress method.
UNIT II LIMIT STATE DESIGN FOR FLEXURE
Analysis and design of singly and doubly reinforced rectangular and flanged beams - nalysis and design of one way, two way and continuous slabs subjected to uniformly istributed load for various boundary conditions.
UNIT III LIMIT STATE DESIGN FOR BOND, ANCHORAGE SHEAR & TORSION
Behaviour of RC members in bond and Anchorage - Design requirements as per current code - Behaviour of RC beams in shear and torsion - Design of RC members for combined bending shear and torsion.
UNIT IV LIMIT STATE DESIGN OF COLUMNS
Types of columns – Braced and unbraced columns – Design of short Rectangular and circular columns for axial, uniaxial and biaxial bending.
UNIT V LIMIT STATE DESIGN OF FOOTING
Design of wall footing – Design of axially and eccentrically loaded rectangular pad and sloped footings – Design of combined rectangular footing for two columns only.
TOTAL: 45 PERIODS
OUTCOMES:
• The student shall be in a position to design the basic elements of reinforced concrete structures.
TEXT BOOKS:
1. Varghese, P.C., “Limit State Design of Reinforced Concrete”, Prentice Hall of India, Pvt. Ltd., New Delhi, 2002.
2. Gambhir.M.L., "Fundamentals of Reinforced Concrete Design", Prentice Hall of India Private Limited, New Delhi, 2006.
3. Subramanian,N.,”Design of Reinforced Concrete Structures”,Oxford University Press, New Delhi, 2013.
REFERENCES:
1. Jain, A.K., “Limit State Design of RC Structures”, Nemchand Publications, Roorkee, 1998
2. Sinha, S.N., “Reinforced Concrete Design”, Tata McGraw Hill Publishing Company Ltd., New Delhi, 2002
3. Unnikrishna Pillai, S., Devdas Menon, “Reinforced Concrete Design”, Tata McGraw Hill Publishing Company Ltd., 2009
4. Punmia.B.C., Ashok Kumar Jain, Arun Kumar Jain, “Limit State Design of Reinforced Concrete”,Laxmi Publication Pvt. Ltd., New Delhi, 2007.
5. Bandyopadhyay. J.N., "Design of Concrete Structures"., Prentice Hall of India Pvt. Ltd., New Delhi, 2008.
6. IS456:2000, Code of practice for Plain and Reinforced Concrete, Bureau of Indian Standards, New Delhi, 2000
7. SP16, IS456:1978 “Design Aids for Reinforced Concrete to Bureau of Indian Standards, New Delhi, 1999
8. Shah V L Karve S R., "Limit State Theory and Design of Reinforced Concrete", Structures Publilcations, Pune, 2013 | 781 | 3,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-35 | longest | en | 0.815857 |
http://publish.illinois.edu/ymb/2017/02/17/solutions-to-2-14/ | 1,558,855,778,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258862.99/warc/CC-MAIN-20190526065059-20190526091059-00156.warc.gz | 166,593,732 | 11,541 | # solutions to 2.14
Exercise:
Let $$V$$ be the space of real polynomial functions of degree at most $$3$$. Consider the quadratic form $$q_1:V \to \mathbb{R}$$ given by $$q_1(f):=|f(-1)|^2+|f(0)|^2+|f(1)|^2.$$ Is this form positive definite?
Consider another form $$q:V\to \mathbb{R}$$ given by $$q(f):=\int_{0}^\infty e^{-x} |f(x)|^2 dx$$ Diagonalize the form using Gram-Schmidt procedure starting with the standard monomial basis $$\{1,x,x^2,x^3\}$$.
Solution:
In order for $$q_1$$ be positive definite, it should vanish only for the zero polynomial, namly $$q_1(f)=0$$ if and only if $$f(x)=0$$. However, the polynomial $$g(x) = x(x-1)(x+1) \in V$$ is nonzero and yet has $$q_1(g) = |g(-1)|^2+|g(0)|^2+|g(1)|2 = 0^2+0^2+0^2 = 0,$$ hence $$q_1$$ is not positive definite on $$V$$.
For the second part, note the bi-linear form corresponding to the quadratic form: $$Q(f,g) = \int_{0}^\infty e^{-x} f(x)g(x) dx$$ and we will utilize the identity $$\int_0^\infty e^{-x} x^n = n!$$ for any integer $$n\geq 1$$.
Starting with the standard basis is $$f_k(x) = x^k$$, the first Gram-Schmidt vector is simply be \begin{align} e_0(x) & = f_0(x) = \boxed{1} \\ \Rightarrow ~~Q(e_0,f_k) &= \int_0^\infty e^{-1} x^k dx = k! \\ Q(e_0,e_0) &= \int_0^\infty e^{-1} 1 dx = 1 \end{align} The second Gram-Schmidt vector follows from that: \begin{align} e_1(x) &= f_1(x) – \frac{Q(e_0,f_1)}{Q(e_0,e_0)}e_0(x) = \boxed{x – 1}, \\ \Rightarrow ~~Q(e_1,e_1) &= \int_0^\infty e^{-1} (x-1)^2 dx = \int_0^\infty e^{-x}(x^2 – 2x +1) dx = 2!-2+1 = 1\\ Q(e_1,f_k) &= \int_0^\infty e^{-1} (x-1)x^k dx = (k+1)!-k! = k\cdot k! \end{align} The third vector: \begin{align} e_2(x) &= f_2(x) – \frac{Q(e_1,f_2)}{Q(e_1,e_1)}e_1(x)- \frac{Q(e_0,f_2)}{Q(e_0,e_0)}e_0(x) = x^2 – 4 (x-1) -2 = \boxed{x^2 -4x + 2}. \\ \Rightarrow ~~Q(e_2,e_2) &= \int_0^\infty e^{-1} (x^2-4x+2)^2 dx = \int_0^\infty e^{-1}(x^4-8x^3 +20x^2 -16x + 4)dx = 4 \\ Q(e_2,f_k) &= \int_0^\infty e^{-1} (x^2-4x+2)x^k dx = (k+2)!-4(k+1)! +2k! \end{align} And the last is: \begin{align} e_3(x) &= f_3(x) – \frac{Q(f_3,e_2)}{Q(e_2,e_2)}e_2(x)- \frac{Q(e_1,f_3)}{Q(e_1,e_1)}e_1(x)- \frac{Q(e_0,f_3)} {Q(e_0,e_0)}e_0(x) \\ &= x^3 – \frac{36}{4}(x^2-4x+2) – \frac{3\cdot 3!}{1}(x-1) – 3! = \boxed{x^3-9x^2 +18x -6} \end{align}
The diagonalized basis $$\{e_0=1,~e_1=x-1,~e_2= x^2-4x+2,~e_3=x^3-9x^2 +18x -6\}$$ is part of the renowned Laguerre polynomials, which are (among other things) orthogonal with respect to the weighted inner product defined by $$Q(f,g)$$ above. That is, for every polynomial $$p(x) = \sum_{k=0}^3 a_k e_k(x)$$ we have $$q(p) = \sum_{k=0}^3 \frac{1}{q(e_k)} a_k^2$$
Exercise: Find the signature of the quadratic form $$q_1(f) =|f(-1)|^2+|f(0)|^2+|f(1)|^2.$$ above.
Solution:
First note that being a sum of non-negative numbers, the form $$q_1(f)\geq 0$$ (i.e no vectors have negative form value) is non-negative definite (sometimes called positive semidefinite). Its signature has $$n_{-} = 0$$ Our next step is to find $$n_0$$ – the dimension of the nullspace of the quadratic form $$q_1$$. If $$q_1(f)=0$$ then it must have three roots at \ (x=\pm 1,0\), hence $$f(x) = h(x) x(x-1)(x+1).$$ However, as we require that $$f\in V$$, or $$\text{deg}(f) \leq 3$$, the factor $$h(x)$$ must be some real constant $$h(x) = \alpha$$, hence all nullvectors of $$q_1(f)=0$$ have the form $$f(x) = \alpha x(x-1)(x+1),$$ forming a one dimensional span, hence $$n_0 = 1$$. Lastly, we compute $$n_{+}$$: since $$\text{dim}(V) = 4 = n_{+} + n_0 + n_{-}$$ we have $$n_{+} = 3$$.
Note: the form $$q_1$$ is positive definite on the space of polynomials of degree $$\leq 2$$ | 1,548 | 3,608 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-22 | latest | en | 0.556243 |
http://onlinetest.ibpsexamguru.in/questions/Clerk-Numerical-Ability/CN-Test-112 | 1,521,753,178,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648003.58/warc/CC-MAIN-20180322205902-20180322225902-00357.warc.gz | 232,532,128 | 8,141 | IBPS Exam Guru
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Home > Bank Clerk > CN Test 112 > General Questions
1 .
A person completes ajourney in 15 hours, the first half at 12 kmh"1 and the second half at 18 kmh"1. What is the total distance ?
212 km 216 km 220 km 224 km
2 .
An 80-metre long train crosses a platform 16 metres long in 12 seconds. What time will it take to cross a bridge of 40m ?
15 sec 20 sec 24 sec 30 sec
3 .
A man takes 4 times as long to row up as to row down the river. If the rate of river is 6 kmh1 , what is the rate of man in still water ?
8 kmh-1 10 kmh-1 12 kmh-1 15 kmh-1
4 .
If the width of a carpet is 5m, how many metres of the carpet will be required to cover a floor of length 40 metres and width 30 metres ?
600 m 420 m 300 m 240 m
5 .
'A' is twrice as fast as 'B' , and is therefore able to finish a work in 24 days less than 'B' . What is the time in which they can do it working together ?
18 days 16 days 20 days 24 days
6 .
A man rode out a certain distance by car at the speed of 45 kmh- 1 and walked back at the rate of 5 kmh1 . The whole journey took 10 hours. What distance did he ride ?
30 km 35 km 40 km 45 km
7 .
The speeds of two trains are in the ratio 2 : 3 . They are going in opposite directions along a parallel line. The first train takes 10 seconds and the second train takes 5 seconds to cross a pole. What is the time they will take to cross each other completely ?
15 sec 7.5 sec 7 sec 6 sec
8 .
A man can row 5 kmh-1 in still water. If the stream is flowing at the speed of 3 kmh-1, it takes him 15 hrs to row to a place and come back. What is the distance between the two places ?
32 km 30 km 28 km 24 km | 547 | 1,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-13 | latest | en | 0.893041 |
https://gbatemp.net/search/648890/ | 1,638,245,317,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358903.73/warc/CC-MAIN-20211130015517-20211130045517-00329.warc.gz | 371,715,099 | 24,249 | # Search results
1. ### DS #5654: May's Mysteries: The Secret of Dragonville (Europe)
If you want to figure out the answer yourself, I can give you some hint: 1. statement 2 is false, so ignore it 2. start off with the "dark blue bottle" by using the hints from statement 1 and 3 to figure out its position Or the answer:
2. ### DS #5654: May's Mysteries: The Secret of Dragonville (Europe)
nevermind, I try out the puzzle again and it is 65795. I probably remembered it wrong, since i finished that puzzle more than a week ago.
3. ### DS #5654: May's Mysteries: The Secret of Dragonville (Europe)
that's weird since my answer is different from your answer but I don't think the answer in the french website (I just went to that website yesterday to look up #78) is right since 5 cannot appear twice I only remember that the answer has two 7s (something like 7*795?), because I misplaced my...
4. ### DS #5654: May's Mysteries: The Secret of Dragonville (Europe)
Okay, I need help again... T-T Puzzle 78 (the bug and bug spray) is driving me crazy..... The instruction says the spray works on 2 horizontal and 2 vertical field, does it mean 1 field on each side of the spray or 2 on each side?? Then "the field with the spray itself mustn't touch a bug"...
5. ### DS #5654: May's Mysteries: The Secret of Dragonville (Europe)
Okay, I think I know which one now, the "divide the inheritance" from the bonus puzzle I took a picture of my solution using my phone, so it is kind of blurry
6. ### DS #5654: May's Mysteries: The Secret of Dragonville (Europe)
THANK YOU, you are a genius! I actually got some of the parts right, but I just couldn't fit chocolate and pistachio into the puzzle, all i know is that the ch in mocha is somehow connected with the ch in pistachio. which one is the drawline inheritance puzzle? what number is it?
7. ### DS #5654: May's Mysteries: The Secret of Dragonville (Europe)
Nevermind, just solved puzzle 50 right after i post the question but I definitely need help on puzzle 48, word puzzle is not my strong field
8. ### DS #5654: May's Mysteries: The Secret of Dragonville (Europe)
anyone reached puzzle 48 or 50 yet?? both of them are very frustrating, 48 is a side puzzle and 50 is required for the story to move on
Help Users
@ KennieDaMeanie: I wouldn't want it for 5 | 602 | 2,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-49 | latest | en | 0.927812 |
https://www.lumoslearning.com/llwp/resources/textbook-resources.html?id=267 | 1,563,311,834,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524879.8/warc/CC-MAIN-20190716201412-20190716223412-00280.warc.gz | 771,068,305 | 19,391 | import_contacts
# SBAC Test Prep: 5th Grade Math Common Core Practice Book and Full-length Online Assessments: Smarter Balanced Study Guide With Performance Task (PT) and Computer Adaptive Testing (CAT)
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### Use the table below to find videos, mobile apps, worksheets and lessons that supplement SBAC Test Prep: 5th Grade Math Common Core Practice Book and Full-length Online Assessments: Smarter Balanced Study Guide With Performance Task (PT) and Computer Adaptive Testing (CAT).
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Report an Error | 477 | 2,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-30 | latest | en | 0.580337 |
https://electronics.stackexchange.com/questions/398273/measuring-current-vs-voltage?noredirect=1 | 1,596,551,652,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.94/warc/CC-MAIN-20200804131928-20200804161928-00339.warc.gz | 284,493,885 | 35,801 | # Measuring current vs voltage
Let's say that I have a periodic charge build-up across the capacitor C1.
I could either choose to measure the current using a transimpedance amplifier as shown below or could measure the voltage instead.
I've been told it's probably better to measure the current because of the parasitic capacitance coming from the BNC cable (represented as a capacitor in the image below).
I do see that the voltage you measure is going to be less for high frequency voltage signal because there will be voltage drop across the parasitic capacitance.
However, don't you also lose current to the parasitic capacitance of the BNC cable? Is the loss miniscule because the parasitic capacitance is on the order of pF? (1st question) So, the parasitic capacitor will get charged up quick and most of the current will still flow to the transimpedance amplifier?
The second question I have is, when using a transimpedance amplifier, as shown below, I was told to put a 1pF capacitor in parallel with the gain resistor to stabilize the feedback.
What is the typical frequency range of the undesired oscillation that we want to get rid of by having this capacitor in parallel with the resistor (to effectively remove the parasitic capacitance of the BNC cable)? Don't you sort of have to know what the unwanted signal's frequency is going to be to choose the right capacitance value to remove that signal? Could someone explain to me what's the best way to choose the right value for the capacitor?
simulate this circuit – Schematic created using CircuitLab
simulate this circuit
The transimpedance amplifier produces a virtual earth at the inverting input and this means it is, in effect, an amplifier with zero ohms input impedance. The 1st knock-on from this is that charge that might have been built up in C1 is immediately converted to current. This is because no charge is allowed to build in C1 because it is effectively shorted by the input circuit's virtual earth. Q = CV and V = 0.
The 2nd knock-on is, that because the input is a short-circuit, it also shorts the parasitic capacitance in the cable and therefore, to a great extent, any effect the parasitic capacitance might have is eradicated.
For your 2nd question the feedback capacitor is not placed there to stabilize the feedback; in fact on many op-amps it can destabilize the circuit. It is there to reduce noise gain at high frequencies; the noise effectively being that produced by the op-amp itself.
See this answer I gave for a better explanation of the problem of noise gain. And here's another relevant answer I gave.
• Thank you for the response. I think everything you said about the benefits of measuring current makes sense. In case we do choose to measure voltage instead though, is the problem the fact that a BNC cable works like a voltage divider (as drawn in the image I just added) and the voltage across the parasitic capacitor ends up becoming the voltage you end up measuring? Thank you! – Blackwidow Sep 27 '18 at 16:53
• When measuring voltage you get a false impression of charge due to the parasitic capacitance taking current proportional to the rate of change of voltage that appears across it. I would recommend using a simulator to get your head around this. – Andy aka Sep 27 '18 at 17:03
• Thank you again. However, if you don't have the resistor that I drew and if the whole thing can be effectively modeled just with a capacitor to the ground, wouldn't the voltage value you measure still be the same as the input voltage? I think without having the voltage divider picture, I don't see how the parasitic capacitance messes up your voltage reading. – Blackwidow Sep 27 '18 at 17:09
• If your voltage measurement is trying to infer charge then adding an unknown capacitor in parallel with C1 means you have an unknown scaling factor. If you know the cable capacitance then measuring voltage is fine. If you measure current you don’t have that unknown scaling factor to deal with nor do you lose sensitivity when the parasitic capacitance is significant. – Andy aka Sep 27 '18 at 17:15 | 895 | 4,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-34 | latest | en | 0.95168 |
https://admin.clutchprep.com/physics/practice-problems/141932/a-biologist-keeps-a-specimen-of-his-favorite-beetle-embedded-in-a-cube-of-polyst | 1,597,514,359,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740929.65/warc/CC-MAIN-20200815154632-20200815184632-00336.warc.gz | 192,415,160 | 20,817 | # Problem: A biologist keeps a specimen of his favorite beetle embedded in a cube of polystyrene plastic. The hapless bug appears to be 2.0 cm within the plastic. What is the beetle's actual distance beneath the surface? Express your answer using two significant figures.
###### FREE Expert Solution
The actual and the apparent depth of the beetle is expressed as:
$\overline{){\mathbit{\eta }}{\mathbf{=}}\frac{{\mathbf{d}}_{\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{u}\mathbf{a}\mathbf{l}}}{{\mathbf{d}}_{\mathbf{a}\mathbf{p}\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{e}\mathbf{n}\mathbf{t}}}}$
Therefore, dactual = ηdapparent
###### Problem Details
A biologist keeps a specimen of his favorite beetle embedded in a cube of polystyrene plastic. The hapless bug appears to be 2.0 cm within the plastic.
What is the beetle's actual distance beneath the surface? Express your answer using two significant figures. | 254 | 910 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-34 | latest | en | 0.741994 |
https://socratic.org/questions/if-25y-4-817-what-is-the-value-of-25y-5#413159 | 1,642,790,404,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303512.46/warc/CC-MAIN-20220121162107-20220121192107-00332.warc.gz | 597,082,092 | 5,771 | # If 25y + 4 = 817 what is the value of 25y + 5?
Apr 26, 2017
See the solution process below:
#### Explanation:
We can write $25 y + 5$ as:
$\left(25 y + 4\right) + 1$
We know $25 y + 4$ is equal to $817$ so we can substitute $817$ for $25 y + 4$ giving:
$\left(817\right) + 1 \implies 817 + 1 \implies 818$ | 124 | 314 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2022-05 | latest | en | 0.569053 |
https://math.stackexchange.com/questions/2413797/how-can-this-illegal-geometry-problem-be-possible/2413812 | 1,656,452,236,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103617931.31/warc/CC-MAIN-20220628203615-20220628233615-00477.warc.gz | 462,714,926 | 62,061 | # How can this "illegal geometry" problem be possible? [duplicate]
Using 2 triangles each with base of 8 and height of 3, and 2 trapezoids with heights of 3 on top, 5 on bottom and height of 5, these four figures can create an area with 64 units squared. However, when rearranged as a rectangle with 13 x 5=65, one additional unit squared seemed to have been created. How is this possible?
• This is the missing square puzzle. Sep 2, 2017 at 0:04
• so the second figure has triangles slightly bigger than the triangles in figure one? Sep 2, 2017 at 0:06
• also, unlike the wikipedia article, the hypothenuse is completely and utterly straight: prntscr.com/gg199b Sep 2, 2017 at 0:10
• @GoodwinLu then it doesn't work, the truth is it's not possible it's an illusion.
– user451844
Sep 2, 2017 at 0:14
• According to the figure on the right, $\frac25=\frac38$. Who knew? Sep 2, 2017 at 1:41
Here's a slightly less subtle "demonstration" that a rectangle with area 5 can be rearranged into a rectangle with area 6.
• Nicely done (+1).
– dxiv
Sep 2, 2017 at 0:47
This is a classic illusion based on the Fibonacci number identity $$13 \times 5 = 1 + 8 \times 8 .$$
The "diagonal" of the rectangle isn't one. The slopes on each segment don't agree. There's one unit of area between the "diagonals".
unlike the wikipedia article, the hypothenuse is completely and utterly straight
No, it's not. Consider the bottom-left corner of the rectangle.
• Let $\alpha$ be the angle in the yellow triangle, then $\tan \alpha = 3/8\,$.
• Let $\beta$ be the angle in the green trapezoid, then $\tan \beta = 5/(5-3)=5/2\,$.
But then $\,\tan \alpha \tan \beta = 15 / 16 \ne 1\,$, so $\,\alpha+\beta \ne 90^\circ\,$ i.e. the two angles do not add up to a right angle. The slopes of the two hypotenuses differ by $\,90^\circ - \arctan 3/8 - \arctan 5/2 \simeq 1.25 ^\circ$.
• In fact, not only is the "hypotenuse" not straight, it is bent in exactly the same way as in the Wikipedia article: part of the "hypotenuse" has slope 2/5, and the rest of it has slope 3/8. Sep 2, 2017 at 0:57
• @DavidK Indeed, and also thanks for the pointer to that cute animation in the duplicate link. Somewhat related: the perpetual chocolate.
– dxiv
Sep 2, 2017 at 1:08 | 691 | 2,239 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-27 | latest | en | 0.907768 |
https://www.answers.com/physics/Express_a_mass_of_0.014_mg_in_grams | 1,709,311,776,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00745.warc.gz | 651,759,036 | 47,160 | 0
# Express a mass of 0.014 mg in grams?
Updated: 8/11/2023
Wiki User
9y ago
1 mg = .001 grams, so .014 mg = 0.000014 grams
Wiki User
9y ago | 58 | 147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.688581 |
https://dspace.vutbr.cz/xmlui/handle/11012/192330?show=full | 1,611,284,223,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529080.43/warc/CC-MAIN-20210122020254-20210122050254-00676.warc.gz | 308,142,379 | 7,228 | Tensors and their applications in mechanics
dc.contributor.advisor Tomáš, Jiří en dc.contributor.author Adejumobi, Mudathir en dc.date.accessioned 2020-07-20T18:58:48Z dc.date.available 2020-07-20T18:58:48Z dc.date.created 2020 cs dc.identifier.citation ADEJUMOBI, M. Tensory a jejich aplikace v mechanice [online]. Brno: Vysoké učení technické v Brně. Fakulta strojního inženýrství. 2020. cs dc.identifier.other 124597 cs dc.identifier.uri http://hdl.handle.net/11012/192330 dc.description.abstract The tensor theory is a branch of Multilinear Algebra that describes the relationship between sets of algebraic objects related to a vector space. Tensor theory together with tensor analysis is usually known to be tensor calculus. This thesis presents a formal category treatment on tensor notation, tensor calculus, and differential manifold. The focus lies mainly on acquiring and understanding the basic concepts of tensors and the operations over them. It looks at how tensor is adapted to differential geometry and continuum mechanics. In particular, it focuses more attention on the application parts of mechanics such as; configuration and deformation, tensor deformation, continuum kinematics, Gauss, and Stokes' theorem with their applications. Finally, it discusses the concept of surface forces and stress vector. en dc.description.abstract The tensor theory is a branch of Multilinear Algebra that describes the relationship between sets of algebraic objects related to a vector space. Tensor theory together with tensor analysis is usually known to be tensor calculus. This thesis presents a formal category treatment on tensor notation, tensor calculus, and differential manifold. The focus lies mainly on acquiring and understanding the basic concepts of tensors and the operations over them. It looks at how tensor is adapted to differential geometry and continuum mechanics. In particular, it focuses more attention on the application parts of mechanics such as; configuration and deformation, tensor deformation, continuum kinematics, Gauss, and Stokes' theorem with their applications. Finally, it discusses the concept of surface forces and stress vector. cs dc.language.iso en cs dc.publisher Vysoké učení technické v Brně. Fakulta strojního inženýrství cs dc.rights Standardní licenční smlouva - přístup k plnému textu bez omezení cs dc.subject Tensors en dc.subject Manifolds en dc.subject Differential manifolds en dc.subject Configuration and deformation en dc.subject Tensor deformation en dc.subject Continuum kinematics en dc.subject Gauss theorem en dc.subject Stokes' theorem en dc.subject Surface forces and stress en dc.subject Tensors cs dc.subject Manifolds cs dc.subject Differential manifolds cs dc.subject Configuration and deformation cs dc.subject Tensor deformation cs dc.subject Continuum kinematics cs dc.subject Gauss theorem cs dc.subject Stokes' theorem cs dc.subject Surface forces and stress cs dc.title Tensory a jejich aplikace v mechanice en dc.title.alternative Tensors and their applications in mechanics cs dc.type Text cs dcterms.dateAccepted 2020-07-16 cs dcterms.modified 2020-07-16-14:29:38 cs thesis.discipline Matematické inženýrství cs thesis.grantor Vysoké učení technické v Brně. Fakulta strojního inženýrství. Ústav matematiky cs thesis.level Inženýrský cs thesis.name Ing. cs sync.item.dbid 124597 en sync.item.dbtype ZP en sync.item.insts 2020.07.20 20:58:48 en sync.item.modts 2020.07.17 08:13:37 en eprints.affiliatedInstitution.faculty Fakulta strojního inženýrství cs dc.contributor.referee Doupovec, Miroslav en dc.description.mark D cs dc.type.driver masterThesis en dc.type.evskp diplomová práce cs but.committee prof. RNDr. Josef Šlapal, CSc. (předseda) prof. RNDr. Miloslav Druckmüller, CSc. (místopředseda) doc. Ing. Luděk Nechvátal, Ph.D. (člen) doc. RNDr. Jiří Tomáš, Dr. (člen) doc. Mgr. Pavel Řehák, Ph.D. (člen) Prof. Bruno Rubino (člen) Assoc. Prof. Matteo Colangeli (člen) Assoc. Prof. Massimiliano Giuli (člen) cs but.defence Student introduced his diploma thesis Tensors and their applications in mechanics to the committee and explained the fundaments of this topic. There was no question in the reviewer's report. Student answered another questions from prof. Šlapal, doc. Tomáš and assoc. prof. Massimiliano Giuli. cs but.result práce byla úspěšně obhájena cs but.program Aplikované vědy v inženýrství cs but.jazyk angličtina (English)
| 1,047 | 4,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-04 | latest | en | 0.823226 |
https://findanyanswer.com/how-do-you-calculate-second-law-efficiency | 1,717,004,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059384.83/warc/CC-MAIN-20240529165728-20240529195728-00174.warc.gz | 212,139,353 | 8,208 | # How do you calculate second law efficiency?
Category: science physics
4.9/5 (180 Views . 30 Votes)
The temperatures here should be in Kelvin → K = ºC + 273.15 or Rankin = 460 + ºF. 2. Second Law Efficiency Second Law efficiency is a measure of how much of the theoretical maximum (Carnot) you achieve, or in other words, a comparison of the system's thermal efficiency to the maximum possible efficiency.
Consequently, what do you mean by second law efficiency?
Exergy efficiency (also known as the second-law efficiency or rational efficiency) computes the effectiveness of a system relative to its performance in reversible conditions. It can also be described as the ratio of the useful work output of the system to the reversible work output for work-consuming systems.
Also, what is First Law efficiency? FIRST LAW EFFICIENCY. The first law states that energy cannot be created or destroyed, but can be converted from one form. to another.
Besides, what is the difference between 1st and 2nd Law Efficiency?
The basic difference between both concepts is that while first law efficiency solely relates to the specific technology applied in order to satisfy certain tasks, for example, illumination and space heating, the concept of second law efficiency takes into account both the actual technology employed and the technology
What is Carnot efficiency?
The Carnot Efficiency is the theoretical maximum efficiency one can get when the heat engine is operating between two temperatures: The temperature at which the high temperature reservoir operates ( THot ).
### What is Exergy destruction?
The exergy destruction of a cycle is the sum of the exergy destruction of the processes that compose that cycle. The exergy destruction of a cycle can also be determined without tracing the individual processes by considering the entire cycle as a single process and using one of the exergy destruction equations.
### What symbol represents internal energy in the first law of thermodynamics?
Then the first law of thermodynamics (ΔU = Q − W) can be used to find the change in internal energy.
### What is irreversibility in thermodynamics?
An irreversible process is a process that cannot return both the system and the surroundings to their original conditions. That is, the system and the surroundings would not return to their original conditions if the process was reversed.
### What is Exergy balance?
Exergy balance consists of internal exergy loss, and represents the results of exergy analysis. In accordance with the second law, the exergy destruction is positive in an irreversible process and vanishes in a reversible process. The change in exergy of a system can be positive, negative, or zero.
### What is refrigerator in thermodynamics?
A refrigerator is a device which is designed to remove heat from a space that is at lower temperature than its surroundings. Effectively any heat engine cycle, when reversed, becomes a refrigeration cycle. The vapor compression cycle is the most commonly used in refrigeration and air condition applications.
### What is the First Law and Second Law of Thermodynamics?
Key Points. The first law, also known as Law of Conservation of Energy, states that energy cannot be created or destroyed in an isolated system. The second law of thermodynamics states that the entropy of any isolated system always increases.
### What does the second law of thermodynamics mean?
The First Law of Thermodynamics states that energy cannot be created or destroyed; the total quantity of energy in the universe stays the same. The Second Law of Thermodynamics is about the quality of energy. It states that as energy is transferred or transformed, more and more of it is wasted.
### What is an example of the second law of thermodynamics?
There are two statements of second law of thermodynamics. Kelvin Plank statement: The best example of this statement is Human Body. We eat food (High temperature reservoir). When we do some work our body warms up and rejects heat into the environment (Low temperature reservoir).
### What is concept of entropy?
Entropy, the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work. Because work is obtained from ordered molecular motion, the amount of entropy is also a measure of the molecular disorder, or randomness, of a system.
### What does the First Law of Thermodynamics mean?
The First Law of Thermodynamics states that heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy. This means that heat energy cannot be created or destroyed.
### What is efficiency in thermodynamics?
Definition of thermodynamic efficiency. : the ratio of work output to heat-energy input in a heat-engine cycle or of heat energy removal to work input in a refrigeration cycle.
### What is Thermodynamics The study of?
Thermodynamics is a branch of physics that deals with heat and temperature, and their relation to energy, work, radiation, and properties of matter. The initial application of thermodynamics to mechanical heat engines was quickly extended to the study of chemical compounds and chemical reactions.
### What is the difference between light and heat?
Heat and light are different but they are both forms of energy. Heat is a form of kinetic energy contained in the random motion of the particles of a material. Light is a form of electromagnetic energy. The higher the temperature of the object, the shorter the wavelength of the EM radiation.
### What is the difference between Exergy and energy?
Exergy is consumed due to irreversibilities. Exergy consumption is proportional to entropy creation. The main important difference between energy and exergy: energy is conserved, while exergy, a measure of energy quality or work potential, can be consumed.
### Is Carnot engine 100 efficient?
No,, never . It can be 100℅ only if the sink temperature is zero or the source temperature is infinite (very large value). Carnot is an ideal cycle is provides benchmark to compare efficiency of other cycles. Carnot efficiency is the maximum efficiency that can be achieved in a cycle.
### What is thermal efficiency of an engine?
The thermal efficiency of a heat engine is the percentage of heat energy that is transformed into work. Thermal efficiency is defined as. The efficiency of even the best heat engines is low; usually below 50% and often far below. | 1,268 | 6,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-22 | latest | en | 0.937009 |
https://proficientwriters.net/buy-essay-online-21548/ | 1,642,908,334,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303956.14/warc/CC-MAIN-20220123015212-20220123045212-00099.warc.gz | 497,966,927 | 8,346 | # Nine fewer than half a number is five more than four times the number. Define a variable write an equation, solve to find the variable and prove your answer.
Nine fewer than half a number is five more than four times the number. Define a variable write an equation, solve to find the variable and prove your answer. | 66 | 318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-05 | latest | en | 0.936451 |
https://free-online-form.com/a-simple-derivation-of-the-centripetal-acceleration-formula/ | 1,656,510,929,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103639050.36/warc/CC-MAIN-20220629115352-20220629145352-00248.warc.gz | 312,355,869 | 12,374 | # A Simple Derivation Of The Centripetal Acceleration Formula
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## Listing Results A Simple Derivation Of The Centripetal Acceleration Formula
### Derivation Of Centripetal Acceleration Detailed
Detailed 49 People Used
5 hours ago The derivation of centripetal acceleration is provided below. Centripetal Acceleration Derivation. The force of a moving object can be written as. From the diagram given above, we can say that, The triangle PQS and AOB are similar. Therefore, Thus, we derive the formula of centripetal acceleration. Students can follow the steps given above to
Category: Centripetal force formula derivation
### A simple derivation of the Centripetal Acceleration Formula?
Simple 60 People Used
Just Now $$\mathbf{a}=\omega \times v \tag{11}$$ And since $$\mathbf{v}=\omega \times r \tag{12}$$ So $$\mathbf{a}=\dfrac {v^{2}}{\mathbf{r}} \tag{13}$$ You can do this derivation by breaking the position of the orbiting particle down into components. It isn't short, but
Category: Formula of centripetal acceleration
### Derivation of Centripetal Acceleration Formula and Proof
Formula 57 People Used
Just Now The centripetal acceleration does not depend on mass .The square of the tangential speed divided by the circle's radius is the centripetal acceleration of an item travelling in a circular route. The force required to maintain the object in that circle varies with mass. A larger mass would necessitate a greater force in the same circular motion.
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### A simple derivation of the Centripetal Acceleration Formula?
Simple 60 People Used
Just Now $\begingroup$ Notice that after one full turn the change in position is also zero. What we are interested in here really the average value of the instantaneous acceleration, but to get it requires calculus (or at least the machinery of limits), which the OP doesn't want.
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### Deriving the centripetal acceleration formula – Matthew
Deriving 58 People Used
5 hours ago Deriving the centripetal acceleration formula. mvaneerde Math, Physics January 24, 2010 October 22, 2020 1 Minute. A body that moves in a circular motion (of radius r) at constant speed (v) is always being accelerated. The acceleration is at right angles to the direction of motion (towards the center of the circle) and of magnitude v 2 / r. The direction of …
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### PhysicsLAB: A Derivation of the Formulas for Centripetal
The 57 People Used
5 hours ago A Derivation of the Formulas for Centripetal Acceleration. Printer Friendly Version : An object is said to be moving in uniform circular motion when it maintains a constant speed while traveling in a circle. Remember that since acceleration is a vector quantity comprised of both magnitude and direction, objects can accelerate in any of these three ways: 1. constant direction, …
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### Centripetal Acceleration Physics Lumen Learning
Physics 49 People Used
9 hours ago Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity v and has the magnitude ac = v2 r;ac =rω2 a c = v 2 r; a c = r ω 2. The unit of centripetal acceleration is m/s 2.
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### Centripetal Acceleration a very easy derivation YouTube
Very 56 People Used
3 hours ago Centripetal acceleration comes into play when a body or particle is moving in a circular path. It is always directed towards the centre of rotation.Sorry for
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### A Derivation of Centripetal Force Cupcake Physics
Force 50 People Used
3 hours ago But then every once in awhile, a young student asks you a seemingly simple question that completely takes you by surprise. That happened to me this past week. Twice! Two undergraduate students came to me on two separate occasions and mentioned that most graduate students and professors could not explain the origin of the equation for centripetal …
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### A simple derivation of the Centripetal Acceleration Formula?
Simple 60 People Used
9 hours ago Home; Physics; A simple derivation of the Centripetal Acceleration Formula? A simple derivation of the Centripetal Acceleration Formula?
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### Centripetal Acceleration Derivation
Simple 35 People Used
2 hours ago We derive both the direction and the equation for centripetal acceleration. This is an AP Physics C: Mechanics topic. Content Times: 0:00 Introduction 1:02 Where centripetal acceleration comes from 4:36 Deriving the Direction of Centripetal Acceleration 8:46 Deriving the Equation for Centripetal Acceleration Thank you to Mrs. Zeiler and the rest of my …
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### Formal Derivation of Centripetal Acceleration—C.E. Mungan
Formal 60 People Used
8 hours ago Formal Derivation of Centripetal Acceleration—C.E. Mungan, Fall 2001 Consider a particle executing uniform circular motion (UCM). Place the origin at the center of the circular trajectory of radius r. The fact that the speed υ is constant means that the angle θ that the position vector r makes with the x-axis increases linearly with time, θ∝t. For reasons that will be discussed in …
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### Centripetal Acceleration Introduction, Example, Formula
Formula 57 People Used
6 hours ago Derivation Of Centripetal Acceleration Formula. The following diagram represents a body moving in a circular pathway at a persistent speed. Here, centripetal acceleration direction is pointed towards the centre of curvature. Furthermore, you must note that triangles created by radii r, Δs and velocity vectors are the same. Additionally, triangles PQR and ABC have two …
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### What is centripetal acceleration Derive the expression
What 55 People Used
3 hours ago Verified. Hint: In circular motion centripetal force is perpendicular to velocity. Also if the particles decrease or increase their speeds in circular motion then acceleration is generated which deviates the net acceleration from pointing towards the centre. We know that the motions are two types: one is straight motion and the circular motion.
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### Full Derivation of the Centripetal Acceleration (No short
Full 58 People Used
1 hours ago meaning that the direction is either out from the center or in towards the center. The negative means the direction is inwards instead of outwards. This proves that the direction of acceleration is always towards the center of the circle. Moving back from polar to cartesian: Just like, . Where is the tangential velocity. Which is the exact equation for centripetal acceleration. Again, …
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### Derive an expression for centripetal acceleration in
Derive 53 People Used
4 hours ago A simple derivation of the Centripetal Acceleration Formula . View Lab Report - Circular Motion.doc from PHY 231 at Northern Virginia Community College. Circular Motion Lab The centripetal force acting on the rotating mass is given by: F 4 2 MR T2 where M PhysicsLAB: 2010 C3. A skier of mass m will be pulled up a hill by a rope, as shown above
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### Derivation of Centripetal Acceleration: Overview
Overview 49 People Used
6 hours ago Illustrated Examples. Example 1) Illustrate the formula of the centripetal acceleration. Answer – The derivation of the centripetal acceleration can be written as a = v2 / r. Example 2) Write the equation of the force of a moving object. Answer – the force of a moving object is F = ma, where F is the force of a moving object, m is the mass
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## New Forms Template
### What is the formula for centripetal acceleration?
Centripetal Acceleration Formula Proof. Consider a particle traversing a circular path of radius r with center O. Initially particle is at P with linear velocity v and angular velocity ω. Since, v = r x ω …(a) Differentiating both the sides w.r.t: dv/dt = ω dr/dt + r dω/dt…(b)
### Why is the derivation of centripetal acceleration important?
The derivation of centripetal acceleration is very important for students who want to learn the concept in-depth. The direction of the centripetal force is towards the center, which is perpendicular to the velocity of the body. The centripetal acceleration derivation will help students to retain the concept for a longer period of time.
### How do you derive the expression for centripetal force?
The centripetal force (F) acting on a particle moving uniformly in a circle depends upon its mass (m), velocity (v) and radius of circle (r). derive the expression for centripetal force using method of dimensions._ (S In physics, you can apply Newton's first and second laws to calculate the centripetal acceleration of an orbiting object.
### When is centripetal acceleration zero in circular motion?
Since, In case of uniform circular motion, the object moves with a constant speed (v), Thus in a circular motion, only centripetal acceleration acts on the body which is given by, This expression can be zeroed when ω = 0. This is possible only when a particle moves in a straight line. | 2,043 | 9,245 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 3, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-27 | latest | en | 0.796286 |
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# Posts Tagged ‘clifford algebra’
## Stokes theorem in Geometric algebra
Posted by peeterjoot on May 17, 2014
[Click here for a PDF of this post with nicer formattingĀ (especially since my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]
Understanding how to apply Stokes theorem to higher dimensional spaces, non-Euclidean metrics, and with curvilinear coordinates has been a long standing goal.
A traditional answer to these questions can be found in the formalism of differential forms, as covered for example in [2], and [8]. However, both of those texts, despite their small size, are intensely scary. I also found it counter intuitive to have to express all physical quantities as forms, since there are many times when we don’t have any pressing desire to integrate these.
Later I encountered Denker’s straight wire treatment [1], which states that the geometric algebra formulation of Stokes theorem has the form
\begin{aligned}\int_S \nabla \wedge F = \int_{\partial S} F\end{aligned} \hspace{\stretch{1}}(1.0.1)
This is simple enough looking, but there are some important details left out. In particular the grades do not match, so there must be some sort of implied projection or dot product operations too. We also need to understand how to express the hypervolume and hypersurfaces when evaluating these integrals, especially when we want to use curvilinear coordinates.
I’d attempted to puzzle through these details previously. A collection of these attempts, to be removed from my collection of geometric algebra notes, can be found in [4]. I’d recently reviewed all of these and wrote a compact synopsis [5] of all those notes, but in the process of doing so, I realized there was a couple of fundamental problems with the approach I had used.
One detail that was that I failed to understand, was that we have a requirement for treating a infinitesimal region in the proof, then summing over such regions to express the boundary integral. Understanding that the boundary integral form and its dot product are both evaluated only at the end points of the integral region is an important detail that follows from such an argument (as used in proof of Stokes theorem for a 3D Cartesian space in [7].)
I also realized that my previous attempts could only work for the special cases where the dimension of the integration volume also equaled the dimension of the vector space. The key to resolving this issue is the concept of the tangent space, and an understanding of how to express the projection of the gradient onto the tangent space. These concepts are covered thoroughly in [6], which also introduces Stokes theorem as a special case of a more fundamental theorem for integration of geometric algebraic objects. My objective, for now, is still just to understand the generalization of Stokes theorem, and will leave the fundamental theorem of geometric calculus to later.
Now that these details are understood, the purpose of these notes is to detail the Geometric algebra form of Stokes theorem, covering its generalization to higher dimensional spaces and non-Euclidean metrics (i.e. especially those used for special relativity and electromagnetism), and understanding how to properly deal with curvilinear coordinates. This generalization has the form
## Theorem 1. Stokes’ Theorem
For blades $F \in \bigwedge^{s}$, and $m$ volume element $d^k \mathbf{x}, s < k$,
\begin{aligned}\int_V d^k \mathbf{x} \cdot (\boldsymbol{\partial} \wedge F) = \int_{\partial V} d^{k-1} \mathbf{x} \cdot F.\end{aligned}
Here the volume integral is over a $m$ dimensional surface (manifold), $\boldsymbol{\partial}$ is the projection of the gradient onto the tangent space of the manifold, and $\partial V$ indicates integration over the boundary of $V$.
It takes some work to give this more concrete meaning. I will attempt to do so in a gradual fashion, and provide a number of examples that illustrate some of the relevant details.
# Basic notation
A finite vector space, not necessarily Euclidean, with basis $\left\{ {\mathbf{e}_1, \mathbf{e}_2, \cdots} \right\}$ will be assumed to be the generator of the geometric algebra. A dual or reciprocal basis $\left\{ {\mathbf{e}^1, \mathbf{e}^2, \cdots} \right\}$ for this basis can be calculated, defined by the property
\begin{aligned}\mathbf{e}_i \cdot \mathbf{e}^j = {\delta_i}^j.\end{aligned} \hspace{\stretch{1}}(1.1.2)
This is an Euclidean space when $\mathbf{e}_i = \mathbf{e}^i, \forall i$.
To select from a multivector $A$ the grade $k$ portion, say $A_k$ we write
\begin{aligned}A_k = {\left\langle A \right\rangle}_{k}.\end{aligned} \hspace{\stretch{1}}(1.1.3)
The scalar portion of a multivector $A$ will be written as
\begin{aligned}{\left\langle A \right\rangle}_{0} \equiv \left\langle A \right\rangle.\end{aligned} \hspace{\stretch{1}}(1.1.4)
The grade selection operators can be used to define the outer and inner products. For blades $U$, and $V$ of grade $r$ and $s$ respectively, these are
\begin{aligned}{\left\langle U V \right\rangle}_{{\left\lvert {r + s} \right\rvert}} \equiv U \wedge V\end{aligned} \hspace{\stretch{1}}(1.0.5.5)
\begin{aligned}{\left\langle U V \right\rangle}_{{\left\lvert {r - s} \right\rvert}} \equiv U \cdot V.\end{aligned} \hspace{\stretch{1}}(1.0.5.5)
Written out explicitly for odd grade blades $A$ (vector, trivector, …), and vector $\mathbf{a}$ the dot and wedge products are respectively
\begin{aligned}\begin{aligned}\mathbf{a} \wedge A &= \frac{1}{2} (\mathbf{a} A - A \mathbf{a}) \\ \mathbf{a} \cdot A &= \frac{1}{2} (\mathbf{a} A + A \mathbf{a}).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.6)
\begin{aligned}\begin{aligned}\mathbf{a} \wedge A &= \frac{1}{2} (\mathbf{a} A + A \mathbf{a}) \\ \mathbf{a} \cdot A &= \frac{1}{2} (\mathbf{a} A - A \mathbf{a}).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.7)
It will be useful to employ the cyclic scalar reordering identity for the scalar selection operator
\begin{aligned}\left\langle{{\mathbf{a} \mathbf{b} \mathbf{c}}}\right\rangle= \left\langle{{\mathbf{b} \mathbf{c} \mathbf{a}}}\right\rangle= \left\langle{{\mathbf{c} \mathbf{a} \mathbf{b}}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.0.8)
For an $N$ dimensional vector space, a product of $N$ orthonormal (up to a sign) unit vectors is referred to as a pseudoscalar for the space, typically denoted by $I$
\begin{aligned}I = \mathbf{e}_1 \mathbf{e}_2 \cdots \mathbf{e}_N.\end{aligned} \hspace{\stretch{1}}(1.0.9)
The pseudoscalar may commute or anticommute with other blades in the space. We may also form a pseudoscalar for a subspace spanned by vectors $\left\{ {\mathbf{a}, \mathbf{b}, \cdots, \mathbf{c}} \right\}$ by unit scaling the wedge products of those vectors $\mathbf{a} \wedge \mathbf{b} \wedge \cdots \wedge \mathbf{c}$.
# Curvilinear coordinates
For our purposes a manifold can be loosely defined as a parameterized surface. For example, a 2D manifold can be considered a surface in an $n$ dimensional vector space, parameterized by two variables
\begin{aligned}\mathbf{x} = \mathbf{x}(a,b) = \mathbf{x}(u^1, u^2).\end{aligned} \hspace{\stretch{1}}(1.0.10)
Note that the indices here do not represent exponentiation. We can construct a basis for the manifold as
\begin{aligned}\mathbf{x}_i = \frac{\partial {\mathbf{x}}}{\partial {u^i}}.\end{aligned} \hspace{\stretch{1}}(1.0.11)
On the manifold we can calculate a reciprocal basis $\left\{ {\mathbf{x}^i} \right\}$, defined by requiring, at each point on the surface
\begin{aligned}\mathbf{x}^i \cdot \mathbf{x}_j = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(1.0.12)
Associated implicitly with this basis is a curvilinear coordinate representation defined by the projection operation
\begin{aligned}\mathbf{x} = x^i \mathbf{x}_i,\end{aligned} \hspace{\stretch{1}}(1.0.13)
(sums over mixed indices are implied). These coordinates can be calculated by taking dot products with the reciprocal frame vectors
\begin{aligned}\mathbf{x} \cdot \mathbf{x}^i &= x^j \mathbf{x}_j \cdot \mathbf{x}^i \\ &= x^j {\delta_j}^i \\ &= x^i.\end{aligned} \hspace{\stretch{1}}(1.0.13)
In this document all coordinates are with respect to a specific curvilinear basis, and not with respect to the standard basis $\left\{ {\mathbf{e}_i} \right\}$ or its dual basis unless otherwise noted.
Similar to the usual notation for derivatives with respect to the standard basis coordinates we form a lower index partial derivative operator
\begin{aligned}\frac{\partial {}}{\partial {u^i}} \equiv \partial_i,\end{aligned} \hspace{\stretch{1}}(1.0.13)
so that when the complete vector space is spanned by $\left\{ {\mathbf{x}_i} \right\}$ the gradient has the curvilinear representation
\begin{aligned}\boldsymbol{\nabla} = \mathbf{x}^i \frac{\partial {}}{\partial {u^i}}.\end{aligned} \hspace{\stretch{1}}(1.0.13)
This can be motivated by noting that the directional derivative is defined by
\begin{aligned}\mathbf{a} \cdot \boldsymbol{\nabla} f(\mathbf{x}) = \lim_{t \rightarrow 0} \frac{f(\mathbf{x} + t \mathbf{a}) - f(\mathbf{x})}{t}.\end{aligned} \hspace{\stretch{1}}(1.0.17)
When the basis $\left\{ {\mathbf{x}_i} \right\}$ does not span the space, the projection of the gradient onto the tangent space at the point of evaluation
\begin{aligned}\boldsymbol{\partial} = \mathbf{x}^i \partial_i = \sum_i \mathbf{x}_i \frac{\partial {}}{\partial {u^i}}.\end{aligned} \hspace{\stretch{1}}(1.0.18)
This is called the vector derivative.
See [6] for a more complete discussion of the gradient and vector derivatives in curvilinear coordinates.
# Green’s theorem
Given a two parameter ($u,v$) surface parameterization, the curvilinear coordinate representation of a vector $\mathbf{f}$ has the form
\begin{aligned}\mathbf{f} = f_u \mathbf{x}^u + f_v \mathbf{x}^v + f_\perp \mathbf{x}^\perp.\end{aligned} \hspace{\stretch{1}}(1.19)
We assume that the vector space is of dimension two or greater but otherwise unrestricted, and need not have an Euclidean basis. Here $f_\perp \mathbf{x}^\perp$ denotes the rejection of $\mathbf{f}$ from the tangent space at the point of evaluation. Green’s theorem relates the integral around a closed curve to an “area” integral on that surface
## Theorem 2. Green’s Theorem
\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} \mathbf{f} \cdot d\mathbf{l}=\iint \left( {-\frac{\partial {f_u}}{\partial {v}}+\frac{\partial {f_v}}{\partial {u}}} \right)du dv\end{aligned}
Following the arguments used in [7] for Stokes theorem in three dimensions, we first evaluate the loop integral along the differential element of the surface at the point $\mathbf{x}(u_0, v_0)$ evaluated over the range $(du, dv)$, as shown in the infinitesimal loop of fig. 1.1.
Fig 1.1. Infinitesimal loop integral
Over the infinitesimal area, the loop integral decomposes into
\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} \mathbf{f} \cdot d\mathbf{l}=\int \mathbf{f} \cdot d\mathbf{x}_1+\int \mathbf{f} \cdot d\mathbf{x}_2+\int \mathbf{f} \cdot d\mathbf{x}_3+\int \mathbf{f} \cdot d\mathbf{x}_4,\end{aligned} \hspace{\stretch{1}}(1.20)
where the differentials along the curve are
\begin{aligned}\begin{aligned}d\mathbf{x}_1 &= {\left.{{ \frac{\partial {\mathbf{x}}}{\partial {u}} }}\right\vert}_{{v = v_0}} du \\ d\mathbf{x}_2 &= {\left.{{ \frac{\partial {\mathbf{x}}}{\partial {v}} }}\right\vert}_{{u = u_0 + du}} dv \\ d\mathbf{x}_3 &= -{\left.{{ \frac{\partial {\mathbf{x}}}{\partial {u}} }}\right\vert}_{{v = v_0 + dv}} du \\ d\mathbf{x}_4 &= -{\left.{{ \frac{\partial {\mathbf{x}}}{\partial {v}} }}\right\vert}_{{u = u_0}} dv.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.21)
It is assumed that the parameterization change $(du, dv)$ is small enough that this loop integral can be considered planar (regardless of the dimension of the vector space). Making use of the fact that $\mathbf{x}^\perp \cdot \mathbf{x}_\alpha = 0$ for $\alpha \in \left\{ {u,v} \right\}$, the loop integral is
\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} \mathbf{f} \cdot d\mathbf{l}=\int\left( {f_u \mathbf{x}^u + f_v \mathbf{x}^v + f_\perp \mathbf{x}^\perp} \right)\cdot\Bigl(\mathbf{x}_u(u, v_0) du - \mathbf{x}_u(u, v_0 + dv) du+\mathbf{x}_v(u_0 + du, v) dv - \mathbf{x}_v(u_0, v) dv\Bigr)=\int f_u(u, v_0) du - f_u(u, v_0 + dv) du+f_v(u_0 + du, v) dv - f_v(u_0, v) dv\end{aligned} \hspace{\stretch{1}}(1.22)
With the distances being infinitesimal, these differences can be rewritten as partial differentials
\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} \mathbf{f} \cdot d\mathbf{l}=\iint \left( {-\frac{\partial {f_u}}{\partial {v}}+\frac{\partial {f_v}}{\partial {u}}} \right)du dv.\end{aligned} \hspace{\stretch{1}}(1.23)
We can now sum over a larger area as in fig. 1.2
Fig 1.2. Sum of infinitesimal loops
All the opposing oriented loop elements cancel, so the integral around the complete boundary of the surface $\mathbf{x}(u, v)$ is given by the $u,v$ area integral of the partials difference.
We will see that Green’s theorem is a special case of the Curl (Stokes) theorem. This observation will also provide a geometric interpretation of the right hand side area integral of thm. 2, and allow for a coordinate free representation.
Special case:
An important special case of Green’s theorem is for a Euclidean two dimensional space where the vector function is
\begin{aligned}\mathbf{f} = P \mathbf{e}_1 + Q \mathbf{e}_2.\end{aligned} \hspace{\stretch{1}}(1.24)
Here Green’s theorem takes the form
\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} P dx + Q dy=\iint \left( {\frac{\partial {Q}}{\partial {x}}-\frac{\partial {P}}{\partial {y}}} \right)dx dy.\end{aligned} \hspace{\stretch{1}}(1.0.25)
# Curl theorem, two volume vector field
Having examined the right hand side of thm. 1 for the very simplest geometric object $\mathbf{f}$, let’s look at the right hand side, the area integral in more detail. We restrict our attention for now to vectors $\mathbf{f}$ still defined by eq. 1.19.
First we need to assign a meaning to $d^2 \mathbf{x}$. By this, we mean the wedge products of the two differential elements. With
\begin{aligned}d\mathbf{x}_i = du^i \frac{\partial {\mathbf{x}}}{\partial {u^i}} = du^i \mathbf{x}_i,\end{aligned} \hspace{\stretch{1}}(1.26)
that area element is
\begin{aligned}d^2 \mathbf{x}= d\mathbf{x}_1 \wedge d\mathbf{x}_2= du^1 du^2 \mathbf{x}_1 \wedge \mathbf{x}_2.\end{aligned} \hspace{\stretch{1}}(1.0.27)
This is the oriented area element that lies in the tangent plane at the point of evaluation, and has the magnitude of the area of that segment of the surface, as depicted in fig. 1.3.
Fig 1.3. Oriented area element tiling of a surface
Observe that we have no requirement to introduce a normal to the surface to describe the direction of the plane. The wedge product provides the information about the orientation of the place in the space, even when the vector space that our vector lies in has dimension greater than three.
Proceeding with the expansion of the dot product of the area element with the curl, using eq. 1.0.6, eq. 1.0.7, and eq. 1.0.8, and a scalar selection operation, we have
\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= \left\langle{{d^2 \mathbf{x} \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)}}\right\rangle \\ &= \left\langle{{d^2 \mathbf{x}\frac{1}{2}\left( { \stackrel{ \rightarrow }{\boldsymbol{\partial}} \mathbf{f} - \mathbf{f} \stackrel{ \leftarrow }{\boldsymbol{\partial}} } \right)}}\right\rangle \\ &= \frac{1}{2}\left\langle{{d^2 \mathbf{x} \left( { \mathbf{x}^i \left( { \partial_i \mathbf{f}} \right) - \left( {\partial_i \mathbf{f}} \right) \mathbf{x}^i } \right)}}\right\rangle \\ &= \frac{1}{2}\left\langle{{\left( { \partial_i \mathbf{f} } \right) d^2 \mathbf{x} \mathbf{x}^i - \left( { \partial_i \mathbf{f} } \right) \mathbf{x}^i d^2 \mathbf{x}}}\right\rangle \\ &= \left\langle{{\left( { \partial_i \mathbf{f} } \right) \left( { d^2 \mathbf{x} \cdot \mathbf{x}^i } \right)}}\right\rangle \\ &= \partial_i \mathbf{f} \cdot\left( { d^2 \mathbf{x} \cdot \mathbf{x}^i } \right).\end{aligned} \hspace{\stretch{1}}(1.28)
Let’s proceed to expand the inner dot product
\begin{aligned}d^2 \mathbf{x} \cdot \mathbf{x}^i &= du^1 du^2\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \mathbf{x}^i \\ &= du^1 du^2\left( {\mathbf{x}_2 \cdot \mathbf{x}^i \mathbf{x}_1-\mathbf{x}_1 \cdot \mathbf{x}^i \mathbf{x}_2} \right) \\ &= du^1 du^2\left( {{\delta_2}^i \mathbf{x}_1-{\delta_1}^i \mathbf{x}_2} \right).\end{aligned} \hspace{\stretch{1}}(1.29)
The complete curl term is thus
\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=du^1 du^2\left( {\frac{\partial {\mathbf{f}}}{\partial {u^2}} \cdot \mathbf{x}_1-\frac{\partial {\mathbf{f}}}{\partial {u^1}} \cdot \mathbf{x}_2} \right)\end{aligned} \hspace{\stretch{1}}(1.30)
This almost has the form of eq. 1.23, although that is not immediately obvious. Working backwards, using the shorthand $u = u^1, v = u^2$, we can show that this coordinate representation can be eliminated
\begin{aligned}-du dv\left( {\frac{\partial {f_v}}{\partial {u}} -\frac{\partial {f_u}}{\partial {v}}} \right) &= du dv\left( {\frac{\partial {}}{\partial {v}}\left( {\mathbf{f} \cdot \mathbf{x}_u} \right)-\frac{\partial {}}{\partial {u}}\left( {\mathbf{f} \cdot \mathbf{x}_v} \right)} \right) \\ &= du dv\left( {\frac{\partial {\mathbf{f}}}{\partial {v}} \cdot \mathbf{x}_u-\frac{\partial {\mathbf{f}}}{\partial {u}} \cdot \mathbf{x}_v+\mathbf{f} \cdot \left( {\frac{\partial {\mathbf{x}_u}}{\partial {v}}-\frac{\partial {\mathbf{x}_v}}{\partial {u}}} \right)} \right) \\ &= du dv \left( {\frac{\partial {\mathbf{f}}}{\partial {v}} \cdot \mathbf{x}_u-\frac{\partial {\mathbf{f}}}{\partial {u}} \cdot \mathbf{x}_v+\mathbf{f} \cdot \left( {\frac{\partial^2 \mathbf{x}}{\partial v \partial u}-\frac{\partial^2 \mathbf{x}}{\partial u \partial v}} \right)} \right) \\ &= du dv \left( {\frac{\partial {\mathbf{f}}}{\partial {v}} \cdot \mathbf{x}_u-\frac{\partial {\mathbf{f}}}{\partial {u}} \cdot \mathbf{x}_v} \right) \\ &= d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right).\end{aligned} \hspace{\stretch{1}}(1.31)
This relates the two parameter surface integral of the curl to the loop integral over its boundary
\begin{aligned}\int d^2 \mathbf{x} \cdot (\boldsymbol{\partial} \wedge \mathbf{f}) = \mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowleft}}\int} \mathbf{f} \cdot d\mathbf{l}.\end{aligned} \hspace{\stretch{1}}(1.0.32)
This is the very simplest special case of Stokes theorem. When written in the general form of Stokes thm. 1
\begin{aligned}\int_A d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f}} \right)=\int_{\partial A} d^1 \mathbf{x} \cdot \mathbf{f}=\int_{\partial A} \left( { d\mathbf{x}_1 - d\mathbf{x}_2 } \right) \cdot \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.0.33)
we must remember (the $\partial A$ is to remind us of this) that it is implied that both the vector $\mathbf{f}$ and the differential elements are evaluated on the boundaries of the integration ranges respectively. A more exact statement is
\begin{aligned}\int_{\partial A} d^1 \mathbf{x} \cdot \mathbf{f}=\int {\left.{{\mathbf{f} \cdot d\mathbf{x}_1}}\right\vert}_{{\Delta u^2}}-{\left.{{\mathbf{f} \cdot d\mathbf{x}_2}}\right\vert}_{{\Delta u^1}}=\int {\left.{{f_1}}\right\vert}_{{\Delta u^2}} du^1-{\left.{{f_2}}\right\vert}_{{\Delta u^1}} du^2.\end{aligned} \hspace{\stretch{1}}(1.0.34)
Expanded out in full this is
\begin{aligned}\int {\left.{{\mathbf{f} \cdot d\mathbf{x}_1}}\right\vert}_{{u^2(1)}}-{\left.{{\mathbf{f} \cdot d\mathbf{x}_1}}\right\vert}_{{u^2(0)}}+{\left.{{\mathbf{f} \cdot d\mathbf{x}_2}}\right\vert}_{{u^1(0)}}-{\left.{{\mathbf{f} \cdot d\mathbf{x}_2}}\right\vert}_{{u^1(1)}},\end{aligned} \hspace{\stretch{1}}(1.0.35)
which can be cross checked against fig. 1.4 to demonstrate that this specifies a clockwise orientation. For the surface with oriented area $d\mathbf{x}_1 \wedge d\mathbf{x}_2$, the clockwise loop is designated with line elements (1)-(4), we see that the contributions around this loop (in boxes) match eq. 1.0.35.
Fig 1.4. Clockwise loop
## Example: Green’s theorem, a 2D Cartesian parameterization for a Euclidean space
For a Cartesian 2D Euclidean parameterization of a vector field and the integration space, Stokes theorem should be equivalent to Green’s theorem eq. 1.0.25. Let’s expand both sides of eq. 1.0.32 independently to verify equality. The parameterization is
\begin{aligned}\mathbf{x}(x, y) = x \mathbf{e}_1 + y \mathbf{e}_2.\end{aligned} \hspace{\stretch{1}}(1.36)
Here the dual basis is the basis, and the projection onto the tangent space is just the gradient
\begin{aligned}\boldsymbol{\partial} = \boldsymbol{\nabla}= \mathbf{e}_1 \frac{\partial {}}{\partial {x}}+ \mathbf{e}_2 \frac{\partial {}}{\partial {y}}.\end{aligned} \hspace{\stretch{1}}(1.0.37)
The volume element is an area weighted pseudoscalar for the space
\begin{aligned}d^2 \mathbf{x} = dx dy \frac{\partial {\mathbf{x}}}{\partial {x}} \wedge \frac{\partial {\mathbf{x}}}{\partial {y}} = dx dy \mathbf{e}_1 \mathbf{e}_2,\end{aligned} \hspace{\stretch{1}}(1.0.38)
and the curl of a vector $\mathbf{f} = f_1 \mathbf{e}_1 + f_2 \mathbf{e}_2$ is
\begin{aligned}\boldsymbol{\partial} \wedge \mathbf{f}=\left( {\mathbf{e}_1 \frac{\partial {}}{\partial {x}}+ \mathbf{e}_2 \frac{\partial {}}{\partial {y}}} \right) \wedge\left( {f_1 \mathbf{e}_1 + f_2 \mathbf{e}_2} \right)=\mathbf{e}_1 \mathbf{e}_2\left( {\frac{\partial {f_2}}{\partial {x}}-\frac{\partial {f_1}}{\partial {y}}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.38)
So, the LHS of Stokes theorem takes the coordinate form
\begin{aligned}\int d^2 \mathbf{x} \cdot (\boldsymbol{\partial} \wedge \mathbf{f}) =\iint dx dy\underbrace{\left\langle{{\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_1 \mathbf{e}_2}}\right\rangle}_{=-1}\left( {\frac{\partial {f_2}}{\partial {x}}-\frac{\partial {f_1}}{\partial {y}}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.38)
For the RHS, following fig. 1.5, we have
\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowleft}}\int} \mathbf{f} \cdot d\mathbf{x}=f_2(x_0, y) dy+f_1(x, y_1) dx-f_2(x_1, y) dy-f_1(x, y_0) dx=\int dx \left( {f_1(x, y_1)-f_1(x, y_0)} \right)-\int dy \left( {f_2(x_1, y)-f_2(x_0, y)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.38)
As expected, we can also obtain this by integrating eq. 1.0.38.
Fig 1.5. Euclidean 2D loop
## Example: Cylindrical parameterization
Let’s now consider a cylindrical parameterization of a 4D space with Euclidean metric $++++$ or Minkowski metric $+++-$. For such a space let’s do a brute force expansion of both sides of Stokes theorem to gain some confidence that all is well.
With $\kappa = \mathbf{e}_3 \mathbf{e}_4$, such a space is conveniently parameterized as illustrated in fig. 1.6 as
\begin{aligned}\mathbf{x}(\rho, \theta, h) = x \mathbf{e}_1 + y \mathbf{e}_2 + \rho \mathbf{e}_3 e^{\kappa \theta}.\end{aligned} \hspace{\stretch{1}}(1.42)
Fig 1.6. Cylindrical polar parameterization
Note that the Euclidean case where $\left( {\mathbf{e}_4} \right)^2 = 1$ rejection of the non-axial components of $\mathbf{x}$ expands to
\begin{aligned}\left( { \left( { \mathbf{x} \wedge \mathbf{e}_1 \wedge \mathbf{e}_2} \right) \cdot \mathbf{e}^2 } \right) \cdot \mathbf{e}^1 =\rho \left( { \mathbf{e}_3 \cos\theta + \mathbf{e}_4 \sin \theta } \right),\end{aligned} \hspace{\stretch{1}}(1.43)
whereas for the Minkowski case where $\left( {\mathbf{e}_4} \right)^2 = -1$ we have a hyperbolic expansion
\begin{aligned}\left( { \left( { \mathbf{x} \wedge \mathbf{e}_1 \wedge \mathbf{e}_2} \right) \cdot \mathbf{e}^2 } \right) \cdot \mathbf{e}^1 =\rho \left( { \mathbf{e}_3 \cosh\theta + \mathbf{e}_4 \sinh \theta } \right).\end{aligned} \hspace{\stretch{1}}(1.44)
Within such a space consider the surface along $x = c, y = d$, for which the vectors are parameterized by
\begin{aligned}\mathbf{x}(\rho, \theta) = c \mathbf{e}_1 + d \mathbf{e}_2 + \rho \mathbf{e}_3 e^{\kappa \theta}.\end{aligned} \hspace{\stretch{1}}(1.45)
The tangent space unit vectors are
\begin{aligned}\mathbf{x}_\rho= \frac{\partial {\mathbf{x}}}{\partial {\rho}} = \mathbf{e}_3 e^{\kappa \theta},\end{aligned} \hspace{\stretch{1}}(1.46)
and
\begin{aligned}\mathbf{x}_\theta &= \frac{\partial {\mathbf{x}}}{\partial {\theta}} \\ &= \rho \mathbf{e}_3 \mathbf{e}_3 \mathbf{e}_4 e^{\kappa \theta} \\ &= \rho \mathbf{e}_4 e^{\kappa \theta}.\end{aligned} \hspace{\stretch{1}}(1.47)
Observe that both of these vectors have their origin at the point of evaluation, and aren’t relative to the absolute origin used to parameterize the complete space.
We wish to compute the volume element for the tangent plane. Noting that $\mathbf{e}_3$ and $\mathbf{e}_4$ both anticommute with $\kappa$ we have for $\mathbf{a} \in \text{span} \left\{ {\mathbf{e}_3, \mathbf{e}_4} \right\}$
\begin{aligned}\mathbf{a} e^{\kappa \theta} = e^{-\kappa \theta} \mathbf{a},\end{aligned} \hspace{\stretch{1}}(1.48)
so
\begin{aligned}\mathbf{x}_\theta \wedge \mathbf{x}_\rho &= {\left\langle{{\mathbf{e}_3 e^{\kappa \theta} \rho \mathbf{e}_4 e^{\kappa \theta}}}\right\rangle}_{2} \\ &= \rho {\left\langle{{\mathbf{e}_3 e^{\kappa \theta} e^{-\kappa \theta} \mathbf{e}_4}}\right\rangle}_{2} \\ &= \rho \mathbf{e}_3 \mathbf{e}_4.\end{aligned} \hspace{\stretch{1}}(1.49)
The tangent space volume element is thus
\begin{aligned}d^2 \mathbf{x} = \rho d\rho d\theta \mathbf{e}_3 \mathbf{e}_4.\end{aligned} \hspace{\stretch{1}}(1.50)
With the tangent plane vectors both perpendicular we don’t need the general lemma 6 to compute the reciprocal basis, but can do so by inspection
\begin{aligned}\mathbf{x}^\rho = e^{-\kappa \theta} \mathbf{e}^3,\end{aligned} \hspace{\stretch{1}}(1.0.51)
and
\begin{aligned}\mathbf{x}^\theta = e^{-\kappa \theta} \mathbf{e}^4 \frac{1}{{\rho}}.\end{aligned} \hspace{\stretch{1}}(1.0.52)
Observe that the latter depends on the metric signature.
The vector derivative, the projection of the gradient on the tangent space, is
\begin{aligned}\boldsymbol{\partial} &= \mathbf{x}^\rho \frac{\partial {}}{\partial {\rho}}+\mathbf{x}^\theta \frac{\partial {}}{\partial {\theta}} \\ &= e^{-\kappa \theta} \left( {\mathbf{e}^3 \partial_\rho + \frac{\mathbf{e}^4}{\rho} \partial_\theta } \right).\end{aligned} \hspace{\stretch{1}}(1.0.52)
From this we see that acting with the vector derivative on a scalar radial only dependent function $f(\rho)$ is a vector function that has a radial direction, whereas the action of the vector derivative on an azimuthal only dependent function $g(\theta)$ is a vector function that has only an azimuthal direction. The interpretation of the geometric product action of the vector derivative on a vector function is not as simple since the product will be a multivector.
Expanding the curl in coordinates is messier, but yields in the end when tackled with sufficient care
\begin{aligned}\boldsymbol{\partial} \wedge \mathbf{f} &= {\left\langle{{e^{-\kappa \theta}\left( { e^3 \partial_\rho + \frac{e^4}{\rho} \partial_\theta} \right)\left( { \not{{e_1 x}} + \not{{e_2 y}} + e_3 e^{\kappa \theta } f_\rho + \frac{e^4}{\rho} e^{\kappa \theta } f_\theta} \right)}}\right\rangle}_{2} \\ &= \not{{{\left\langle{{e^{-\kappa \theta} e^3 \partial_\rho \left( { e_3 e^{\kappa \theta } f_\rho} \right)}}\right\rangle}_{2}}}+{\left\langle{{\not{{e^{-\kappa \theta}}} e^3 \partial_\rho \left( { \frac{e^4}{\rho} \not{{e^{\kappa \theta }}} f_\theta} \right)}}\right\rangle}_{2}+{\left\langle{{e^{-\kappa \theta}\frac{e^4}{\rho} \partial_\theta\left( { e_3 e^{\kappa \theta } f_\rho} \right)}}\right\rangle}_{2}+{\left\langle{{e^{-\kappa \theta}\frac{e^4}{\rho} \partial_\theta\left( { \frac{e^4}{\rho} e^{\kappa \theta } f_\theta} \right)}}\right\rangle}_{2} \\ &= \mathbf{e}^3 \mathbf{e}^4 \left( {-\frac{f_\theta}{\rho^2} + \frac{1}{{\rho}} \partial_\rho f_\theta- \frac{1}{{\rho}} \partial_\theta f_\rho} \right)+ \frac{1}{{\rho^2}}{\left\langle{{e^{-\kappa \theta} \left( {\mathbf{e}^4} \right)^2\left( {\mathbf{e}_3 \mathbf{e}_4 f_\theta+ \not{{\partial_\theta f_\theta}}} \right)e^{\kappa \theta}}}\right\rangle}_{2} \\ &= \mathbf{e}^3 \mathbf{e}^4 \left( {-\frac{f_\theta}{\rho^2} + \frac{1}{{\rho}} \partial_\rho f_\theta- \frac{1}{{\rho}} \partial_\theta f_\rho} \right)+ \frac{1}{{\rho^2}}{\left\langle{{\not{{e^{-\kappa \theta} }}\mathbf{e}_3 \mathbf{e}^4 f_\theta\not{{e^{\kappa \theta}}}}}\right\rangle}_{2} \\ &= \frac{\mathbf{e}^3 \mathbf{e}^4 }{\rho}\left( {\partial_\rho f_\theta- \partial_\theta f_\rho} \right).\end{aligned} \hspace{\stretch{1}}(1.0.52)
After all this reduction, we can now state in coordinates the LHS of Stokes theorem explicitly
\begin{aligned}\int d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= \int \rho d\rho d\theta \left\langle{{\mathbf{e}_3 \mathbf{e}_4 \mathbf{e}^3 \mathbf{e}^4 }}\right\rangle\frac{1}{{\rho}}\left( {\partial_\rho f_\theta- \partial_\theta f_\rho} \right) \\ &= \int d\rho d\theta\left( {\partial_\theta f_\rho-\partial_\rho f_\theta} \right) \\ &= \int d\rho {\left.{{f_\rho}}\right\vert}_{{\Delta \theta}}- \int d\theta{\left.{{f_\theta}}\right\vert}_{{\Delta \rho}}.\end{aligned} \hspace{\stretch{1}}(1.0.52)
Now compare this to the direct evaluation of the loop integral portion of Stokes theorem. Expressing this using eq. 1.0.34, we have the same result
\begin{aligned}\int d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=\int {\left.{{f_\rho}}\right\vert}_{{\Delta \theta}} d\rho-{\left.{{f_\theta}}\right\vert}_{{\Delta \rho}} d\theta\end{aligned} \hspace{\stretch{1}}(1.0.56)
This example highlights some of the power of Stokes theorem, since the reduction of the volume element differential form was seen to be quite a chore (and easy to make mistakes doing.)
## Example: Composition of boost and rotation
Working in a $\bigwedge^{1,3}$ space with basis $\left\{ {\gamma_0, \gamma_1, \gamma_2, \gamma_3} \right\}$ where $\left( {\gamma_0} \right)^2 = 1$ and $\left( {\gamma_k} \right)^2 = -1, k \in \left\{ {1,2,3} \right\}$, an active composition of boost and rotation has the form
\begin{aligned}\begin{aligned}\mathbf{x}' &= e^{i\alpha/2} \mathbf{x}_0 e^{-i\alpha/2} \\ \mathbf{x}'' &= e^{-j\theta/2} \mathbf{x}' e^{j\theta/2}\end{aligned},\end{aligned} \hspace{\stretch{1}}(1.0.57)
where $i$ is a bivector of a timelike unit vector and perpendicular spacelike unit vector, and $j$ is a bivector of two perpendicular spacelike unit vectors. For example, $i = \gamma_0 \gamma_1$ and $j = \gamma_1 \gamma_2$. For such $i,j$ the respective Lorentz transformation matrices are
\begin{aligned}{\begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}}'=\begin{bmatrix}\cosh\alpha & -\sinh\alpha & 0 & 0 \\ -\sinh\alpha & \cosh\alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.58)
and
\begin{aligned}{\begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}}''=\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & \cos\theta & \sin\theta & 0 \\ 0 & -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}{\begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}}'.\end{aligned} \hspace{\stretch{1}}(1.0.59)
Let’s calculate the tangent space vectors for this parameterization, assuming that the particle is at an initial spacetime position of $\mathbf{x}_0$. That is
\begin{aligned}\mathbf{x} = e^{-j\theta/2} e^{i\alpha/2} \mathbf{x}_0e^{-i\alpha/2} e^{j\theta/2}.\end{aligned} \hspace{\stretch{1}}(1.0.60)
To calculate the tangent space vectors for this subspace we note that
\begin{aligned}\frac{\partial {\mathbf{x}'}}{\partial {\alpha}} = \frac{i}{2} \mathbf{x}_0 - \mathbf{x}_0 \frac{i}{2} = i \cdot \mathbf{x}_0,\end{aligned} \hspace{\stretch{1}}(1.0.61)
and
\begin{aligned}\frac{\partial {\mathbf{x}''}}{\partial {\theta}} = -\frac{j}{2} \mathbf{x}' + \mathbf{x}' \frac{j}{2} = \mathbf{x}' \cdot j.\end{aligned} \hspace{\stretch{1}}(1.0.62)
The tangent space vectors are therefore
\begin{aligned}\begin{aligned}\mathbf{x}_\alpha &= e^{-j\theta/2} \left( { i \cdot \mathbf{x}_0 } \right)e^{j\theta/2} \\ \mathbf{x}_\theta &= \left( {e^{i\alpha/2} \mathbf{x}_0e^{-i\alpha/2} } \right) \cdot j.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.63)
Continuing a specific example where $i = \gamma_0\gamma_1, j = \gamma_1 \gamma_2$ let’s also pick $\mathbf{x}_0 = \gamma_0$, the spacetime position of a particle at the origin of a frame at that frame’s $c t = 1$. The tangent space vectors for the subspace parameterized by this transformation and this initial position is then reduced to
\begin{aligned}\mathbf{x}_\alpha = -\gamma_1 e^{j \theta} = \gamma_1 \sin\theta + \gamma_2 \cos\theta,\end{aligned} \hspace{\stretch{1}}(1.0.63)
and
\begin{aligned}\mathbf{x}_\theta &= \left( { \gamma_0 e^{-i \alpha} } \right) \cdot j \\ &= \left( { \gamma_0\left( { \cosh\alpha - \gamma_0 \gamma_1 \sinh\alpha } \right)} \right) \cdot \left( { \gamma_1 \gamma_2} \right) \\ &= {\left\langle{{ \left( { \gamma_0 \cosh\alpha - \gamma_1 \sinh\alpha } \right) \gamma_1 \gamma_2 }}\right\rangle}_{1} \\ &= \gamma_2 \sinh\alpha.\end{aligned} \hspace{\stretch{1}}(1.0.63)
By inspection the dual basis for this parameterization is
\begin{aligned}\begin{aligned}\mathbf{x}^\alpha &= \gamma_1 e^{j \theta} \\ \mathbf{x}^\theta &= \frac{\gamma^2}{\sinh\alpha} \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.66)
So, Stokes theorem, applied to a spacetime vector $\mathbf{f}$, for this subspace is
\begin{aligned}\int d\alpha d\theta \sinh\alpha \sin\theta \left( { \gamma_1 \gamma_2 } \right) \cdot \left( {\left( {\gamma_1 e^{j \theta} \partial_\alpha + \frac{\gamma^2}{\sinh\alpha} \partial_\theta} \right)\wedge \mathbf{f}} \right)=\int d\alpha {\left.{{\mathbf{f} \cdot \Bigl( {\gamma^1 e^{j \theta}} \Bigr)}}\right\vert}_{{\theta_0}}^{{\theta_1}}-\int d\theta {\left.{{\mathbf{f} \cdot \Bigl( { \gamma_2 \sinh\alpha } \Bigr)}}\right\vert}_{{\alpha_0}}^{{\alpha_1}}.\end{aligned} \hspace{\stretch{1}}(1.0.67)
Since the point is to avoid the curl integral, we did not actually have to state it explicitly, nor was there any actual need to calculate the dual basis.
## Example: Dual representation in three dimensions
It’s clear that there is a projective nature to the differential form $d^2 \mathbf{x} \cdot \left( {\boldsymbol{\partial} \wedge \mathbf{f}} \right)$. This projective nature allows us, in three dimensions, to re-express Stokes theorem using the gradient instead of the vector derivative, and to utilize the cross product and a normal direction to the plane.
When we parameterize a normal direction to the tangent space, so that for a 2D tangent space spanned by curvilinear coordinates $\mathbf{x}_1$ and $\mathbf{x}_2$ the vector $\mathbf{x}^3$ is normal to both, we can write our vector as
\begin{aligned}\mathbf{f} = f_1 \mathbf{x}^1 + f_2 \mathbf{x}^2 + f_3 \mathbf{x}^3,\end{aligned} \hspace{\stretch{1}}(1.0.68)
and express the orientation of the tangent space area element in terms of a pseudoscalar that includes this normal direction
\begin{aligned}\mathbf{x}_1 \wedge \mathbf{x}_2 =\mathbf{x}^3 \cdot \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) =\mathbf{x}^3 \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right).\end{aligned} \hspace{\stretch{1}}(1.0.69)
Inserting this into an expansion of the curl form we have
\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= du^1 du^2 \left\langle{{\mathbf{x}^3 \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right)\left( {\left( {\sum_{i=1,2} x^i \partial_i} \right)\wedge\mathbf{f}} \right)}}\right\rangle \\ &= du^1 du^2 \mathbf{x}^3 \cdot \left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right)\cdot \left( {\boldsymbol{\nabla} \wedge \mathbf{f}} \right)-\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right)\cdot \left( {\mathbf{x}^3 \partial_3 \wedge \mathbf{f}} \right)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.69)
Observe that this last term, the contribution of the component of the gradient perpendicular to the tangent space, has no $\mathbf{x}_3$ components
\begin{aligned}\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right)\cdot \left( {\mathbf{x}^3 \partial_3 \wedge \mathbf{f}} \right) &= \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right)\cdot \left( {\mathbf{x}^3 \wedge \partial_3 \mathbf{f}} \right) \\ &= \left( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \mathbf{x}^3} \right)\cdot \partial_3 \mathbf{f} \\ &= \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \partial_3 \mathbf{f} \\ &= \mathbf{x}_1 \left( { \mathbf{x}_2 \cdot \partial_3 \mathbf{f} } \right)-\mathbf{x}_2 \left( { \mathbf{x}_1 \cdot \partial_3 \mathbf{f} } \right),\end{aligned} \hspace{\stretch{1}}(1.0.69)
leaving
\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=du^1 du^2 \mathbf{x}^3 \cdot \left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \left( { \boldsymbol{\nabla} \wedge \mathbf{f}} \right)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.69)
Now scale the normal vector and its dual to have unit norm as follows
\begin{aligned}\begin{aligned}\mathbf{x}^3 &= \alpha \hat{\mathbf{x}}^3 \\ \mathbf{x}_3 &= \frac{1}{{\alpha}} \hat{\mathbf{x}}_3,\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.73)
so that for $\beta > 0$, the volume element can be
\begin{aligned}\mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \hat{\mathbf{x}}_3 = \beta I.\end{aligned} \hspace{\stretch{1}}(1.0.73)
This scaling choice is illustrated in fig. 1.7, and represents the “outwards” normal. With such a scaling choice we have
Fig 1.7. Outwards normal
\begin{aligned}\beta du^1 du^2 = dA,\end{aligned} \hspace{\stretch{1}}(1.75)
and almost have the desired cross product representation
\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=dA \hat{\mathbf{x}}^3 \cdot \left( { I \cdot \left( {\boldsymbol{\nabla} \wedge \mathbf{f}} \right) } \right)=dA \hat{\mathbf{x}}^3 \cdot \left( { I \left( {\boldsymbol{\nabla} \wedge \mathbf{f}} \right) } \right).\end{aligned} \hspace{\stretch{1}}(1.76)
With the duality identity $\mathbf{a} \wedge \mathbf{b} = I \left( {\mathbf{a} \times \mathbf{b}} \right)$, we have the traditional 3D representation of Stokes theorem
\begin{aligned}\int d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=-\int dA \hat{\mathbf{x}}^3 \cdot \left( {\boldsymbol{\nabla} \times \mathbf{f}} \right) = \mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowleft}}\int} \mathbf{f} \cdot d\mathbf{l}.\end{aligned} \hspace{\stretch{1}}(1.0.77)
Note that the orientation of the loop integral in the traditional statement of the 3D Stokes theorem is counterclockwise instead of clockwise, as written here.
# Stokes theorem, three variable volume element parameterization
We can restate the identity of thm. 1 in an equivalent dot product form.
\begin{aligned}\int_V \left( { d^k \mathbf{x} \cdot \mathbf{x}^i } \right) \cdot \partial_i F = \int_{\partial V} d^{k-1} \mathbf{x} \cdot F.\end{aligned} \hspace{\stretch{1}}(1.0.78)
Here $d^{k-1} \mathbf{x} = \sum_i d^k \mathbf{x} \cdot \mathbf{x}^i$, with the implicit assumption that it and the blade $F$ that it is dotted with, are both evaluated at the end points of integration variable $u^i$ that has been integrated against.
We’ve seen one specific example of this above in the expansions of eq. 1.28, and eq. 1.29, however, the equivalent result of eq. 1.0.78, somewhat magically, applies to any degree blade and volume element provided the degree of the blade is less than that of the volume element (i.e. $s < k$). That magic follows directly from lemma 1.
As an expositional example, consider a three variable volume element parameterization, and a vector blade $\mathbf{f}$
\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= \left( { d^3 \mathbf{x} \cdot \mathbf{x}^i } \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3\left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \mathbf{x}^i } \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3\left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) {\delta_3}^i-\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 } \right) {\delta_2}^i+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) {\delta_1}^i} \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3\left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \partial_3 \mathbf{f}-\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 } \right) \cdot \partial_2 \mathbf{f}+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_1 \mathbf{f}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.78)
It should not be surprising that this has the structure found in the theory of differential forms. Using the differentials for each of the parameterization “directions”, we can write this dot product expansion as
\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=\left( {du^3 \left( { d\mathbf{x}_1 \wedge d\mathbf{x}_2 } \right) \cdot \partial_3 \mathbf{f}-du^2 \left( { d\mathbf{x}_1 \wedge d\mathbf{x}_3 } \right) \cdot \partial_2 \mathbf{f}+du^1 \left( { d\mathbf{x}_2 \wedge d\mathbf{x}_3 } \right) \cdot \partial_1 \mathbf{f}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.78)
Observe that the sign changes with each element of $d\mathbf{x}_1 \wedge d\mathbf{x}_2 \wedge d\mathbf{x}_3$ that is skipped. In differential forms, the wedge product composition of 1-forms is an abstract quantity. Here the differentials are just vectors, and their wedge product represents an oriented volume element. This interpretation is likely available in the theory of differential forms too, but is arguably less obvious.
## Digression
As was the case with the loop integral, we expect that the coordinate representation has a representation that can be expressed as a number of antisymmetric terms. A bit of experimentation shows that such a sum, after dropping the parameter space volume element factor, is
\begin{aligned}\mathbf{x}_1 \left( { -\partial_2 f_3 + \partial_3 f_2 } \right)+\mathbf{x}_2 \left( { -\partial_3 f_1 + \partial_1 f_3 } \right)+\mathbf{x}_3 \left( { -\partial_1 f_2 + \partial_2 f_1 } \right) &= \mathbf{x}_1 \left( { -\partial_2 \mathbf{f} \cdot \mathbf{x}_3 + \partial_3 \mathbf{f} \cdot \mathbf{x}_2 } \right)+\mathbf{x}_2 \left( { -\partial_3 \mathbf{f} \cdot \mathbf{x}_1 + \partial_1 \mathbf{f} \cdot \mathbf{x}_3 } \right)+\mathbf{x}_3 \left( { -\partial_1 \mathbf{f} \cdot \mathbf{x}_2 + \partial_2 \mathbf{f} \cdot \mathbf{x}_1 } \right) \\ &= \left( { \mathbf{x}_1 \partial_3 \mathbf{f} \cdot \mathbf{x}_2 -\mathbf{x}_2 \partial_3 \mathbf{f} \cdot \mathbf{x}_1 } \right)+\left( { \mathbf{x}_3 \partial_2 \mathbf{f} \cdot \mathbf{x}_1 -\mathbf{x}_1 \partial_2 \mathbf{f} \cdot \mathbf{x}_3 } \right)+\left( { \mathbf{x}_2 \partial_1 \mathbf{f} \cdot \mathbf{x}_3 -\mathbf{x}_3 \partial_1 \mathbf{f} \cdot \mathbf{x}_2 } \right) \\ &= \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \partial_3 \mathbf{f}+\left( { \mathbf{x}_3 \wedge \mathbf{x}_1 } \right) \cdot \partial_2 \mathbf{f}+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_1 \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.0.78)
To proceed with the integration, we must again consider an infinitesimal volume element, for which the partial can be evaluated as the difference of the endpoints, with all else held constant. For this three variable parameterization, say, $(u,v,w)$, let’s delimit such an infinitesimal volume element by the parameterization ranges $[u_0,u_0 + du]$, $[v_0,v_0 + dv]$, $[w_0,w_0 + dw]$. The integral is
\begin{aligned}\begin{aligned}\int_{u = u_0}^{u_0 + du}\int_{v = v_0}^{v_0 + dv}\int_{w = w_0}^{w_0 + dw}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)&=\int_{u = u_0}^{u_0 + du}du\int_{v = v_0}^{v_0 + dv}dv{\left.{{ \Bigl( { \left( { \mathbf{x}_u \wedge \mathbf{x}_v } \right) \cdot \mathbf{f} } \Bigr) }}\right\vert}_{{w = w_0}}^{{w_0 + dw}} \\ &-\int_{u = u_0}^{u_0 + du}du\int_{w = w_0}^{w_0 + dw}dw{\left.{{\Bigl( { \left( { \mathbf{x}_u \wedge \mathbf{x}_w } \right) \cdot \mathbf{f} } \Bigr) }}\right\vert}_{{v = v_0}}^{{v_0 + dv}} \\ &+\int_{v = v_0}^{v_0 + dv}dv\int_{w = w_0}^{w_0 + dw}dw{\left.{{\Bigl( { \left( { \mathbf{x}_v \wedge \mathbf{x}_w } \right) \cdot \mathbf{f} } \Bigr) }}\right\vert}_{{u = u_0}}^{{u_0 + du}}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.82)
Extending this over the ranges $[u_0,u_0 + \Delta u]$, $[v_0,v_0 + \Delta v]$, $[w_0,w_0 + \Delta w]$, we have proved Stokes thm. 1 for vectors and a three parameter volume element, provided we have a surface element of the form
\begin{aligned}d^2 \mathbf{x} = {\left. \Bigl( {d\mathbf{x}_u \wedge d\mathbf{x}_v } \Bigr) \right\vert}_{w = w_0}^{w_1}-{\left. \Bigl( {d\mathbf{x}_u \wedge d\mathbf{x}_w } \Bigr) \right\vert}_{v = v_0}^{v_1}+{\left. \Bigl( {d\mathbf{x}_v \wedge \mathbf{x}_w } \Bigr) \right\vert}_{ u = u_0 }^{u_1},\end{aligned} \hspace{\stretch{1}}(1.0.82)
where the evaluation of the dot products with $\mathbf{f}$ are also evaluated at the same points.
## Example: Euclidean spherical polar parameterization of 3D subspace
Consider an Euclidean space where a 3D subspace is parameterized using spherical coordinates, as in
\begin{aligned}\mathbf{x}(x, \rho, \theta, \phi) = \mathbf{e}_1 x + \mathbf{e}_4 \rho \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta} \right)=\left( {x, \rho \sin\theta \cos\phi, \rho \sin\theta \sin\phi, \rho \cos\theta} \right).\end{aligned} \hspace{\stretch{1}}(1.0.84)
The tangent space basis for the subspace situated at some fixed $x = x_0$, is easy to calculate, and is found to be
\begin{aligned}\begin{aligned}\mathbf{x}_\rho &= \left( {0, \sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta} \right) =\mathbf{e}_4 \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta} \right) \\ \mathbf{x}_\theta &= \rho \left( {0, \cos\theta \cos\phi, \cos\theta \sin\phi, - \sin\theta} \right) =\rho \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta } \right) \\ \mathbf{x}_\phi &=\rho \left( {0, -\sin\theta \sin\phi, \sin\theta \cos\phi, 0} \right)= \rho \sin\theta \mathbf{e}_3 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.85)
While we can use the general relation of lemma 7 to compute the reciprocal basis. That is
\begin{aligned}\mathbf{a}^{*} = \left( { \mathbf{b} \wedge \mathbf{c} } \right) \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} }}.\end{aligned} \hspace{\stretch{1}}(1.0.86)
However, a naive attempt at applying this without algebraic software is a route that requires a lot of care, and is easy to make mistakes doing. In this case it is really not necessary since the tangent space basis only requires scaling to orthonormalize, satisfying for $i,j \in \left\{ {\rho, \theta, \phi} \right\}$
\begin{aligned}\mathbf{x}_i \cdot \mathbf{x}_j =\begin{bmatrix} 1 & 0 & 0 \\ 0 & \rho^2 & 0 \\ 0 & 0 & \rho^2 \sin^2 \theta \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.87)
This allows us to read off the dual basis for the tangent volume by inspection
\begin{aligned}\begin{aligned}\mathbf{x}^\rho &=\mathbf{e}_4 \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta} \right) \\ \mathbf{x}^\theta &= \frac{1}{{\rho}} \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta } \right) \\ \mathbf{x}^\phi &=\frac{1}{{\rho \sin\theta}} \mathbf{e}_3 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.88)
Should we wish to explicitly calculate the curl on the tangent space, we would need these. The area and volume elements are also messy to calculate manually. This expansion can be found in the Mathematica notebook \nbref{sphericalSurfaceAndVolumeElements.nb}, and is
\begin{aligned}\begin{aligned}\mathbf{x}_\theta \wedge \mathbf{x}_\phi &=\rho^2 \sin\theta \left( \mathbf{e}_4 \mathbf{e}_2 \sin\theta \sin\phi + \mathbf{e}_2 \mathbf{e}_3 \cos\theta + \mathbf{e}_3 \mathbf{e}_4 \sin\theta \cos\phi \right) \\ \mathbf{x}_\phi \wedge \mathbf{x}_\rho &=\rho \sin\theta \left(-\mathbf{e}_2 \mathbf{e}_3 \sin\theta -\mathbf{e}_2 \mathbf{e}_4 \cos\theta \sin\phi +\mathbf{e}_3 \mathbf{e}_4\cos\theta \cos\phi \right) \\ \mathbf{x}_\rho \wedge \mathbf{x}_\theta &= -\mathbf{e}_4 \rho \left(\mathbf{e}_2\cos\phi +\mathbf{e}_3\sin\phi \right) \\ \mathbf{x}_\rho \wedge \mathbf{x}_\theta \wedge \mathbf{x}_\phi &= \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4 \rho^2 \sin\theta \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.89)
Those area elements have a Geometric algebra factorization that are perhaps useful
\begin{aligned}\begin{aligned}\mathbf{x}_\theta \wedge \mathbf{x}_\phi &=-\rho^2 \sin\theta \mathbf{e}_2 \mathbf{e}_3 \exp\left( {-\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta} \right) \\ \mathbf{x}_\phi \wedge \mathbf{x}_\rho &=\rho \sin\theta \mathbf{e}_3 \mathbf{e}_4 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}\exp\left( {\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta} \right) \\ \mathbf{x}_\rho \wedge \mathbf{x}_\theta &= -\rho \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}\end{aligned}.\end{aligned} \hspace{\stretch{1}}(1.0.90)
One of the beauties of Stokes theorem is that we don’t actually have to calculate the dual basis on the tangent space to proceed with the integration. For that calculation above, where we had a normal tangent basis, I still used software was used as an aid, so it is clear that this can generally get pretty messy.
To apply Stokes theorem to a vector field we can use eq. 1.0.82 to write down the integral directly
\begin{aligned}\int_V d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= \int_{\partial V} d^2 \mathbf{x} \cdot \mathbf{f} \\ &= \int {\left.{{ \left( { \mathbf{x}_\theta \wedge \mathbf{x}_\phi } \right) \cdot \mathbf{f} }}\right\vert}_{{\rho = \rho_0}}^{{\rho_1}} d\theta d\phi+\int{\left.{{ \left( { \mathbf{x}_\phi \wedge \mathbf{x}_\rho } \right) \cdot \mathbf{f} }}\right\vert}_{{\theta = \theta_0}}^{{\theta_1}} d\phi d\rho+\int{\left.{{ \left( { \mathbf{x}_\rho \wedge \mathbf{x}_\theta } \right) \cdot \mathbf{f} }}\right\vert}_{{\phi = \phi_0}}^{{\phi_1}} d\rho d\theta.\end{aligned} \hspace{\stretch{1}}(1.0.90)
Observe that eq. 1.0.90 is a vector valued integral that expands to
\begin{aligned}\int {\left.{{ \left( { \mathbf{x}_\theta f_\phi - \mathbf{x}_\phi f_\theta } \right) }}\right\vert}_{{\rho = \rho_0}}^{{\rho_1}} d\theta d\phi+\int {\left.{{ \left( { \mathbf{x}_\phi f_\rho - \mathbf{x}_\rho f_\phi } \right) }}\right\vert}_{{\theta = \theta_0}}^{{\theta_1}} d\phi d\rho+\int {\left.{{ \left( { \mathbf{x}_\rho f_\theta - \mathbf{x}_\theta f_\rho } \right) }}\right\vert}_{{\phi = \phi_0}}^{{\phi_1}} d\rho d\theta.\end{aligned} \hspace{\stretch{1}}(1.0.92)
This could easily be a difficult integral to evaluate since the vectors $\mathbf{x}_i$ evaluated at the endpoints are still functions of two parameters. An easier integral would result from the application of Stokes theorem to a bivector valued field, say $B$, for which we have
\begin{aligned}\int_V d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right) &= \int_{\partial V} d^2 \mathbf{x} \cdot B \\ &= \int {\left.{{ \left( { \mathbf{x}_\theta \wedge \mathbf{x}_\phi } \right) \cdot B }}\right\vert}_{{\rho = \rho_0}}^{{\rho_1}} d\theta d\phi+\int{\left.{{ \left( { \mathbf{x}_\phi \wedge \mathbf{x}_\rho } \right) \cdot B }}\right\vert}_{{\theta = \theta_0}}^{{\theta_1}} d\phi d\rho+\int{\left.{{ \left( { \mathbf{x}_\rho \wedge \mathbf{x}_\theta } \right) \cdot B }}\right\vert}_{{\phi = \phi_0}}^{{\phi_1}} d\rho d\theta \\ &= \int {\left.{{ B_{\phi \theta} }}\right\vert}_{{\rho = \rho_0}}^{{\rho_1}} d\theta d\phi+\int{\left.{{ B_{\rho \phi} }}\right\vert}_{{\theta = \theta_0}}^{{\theta_1}} d\phi d\rho+\int{\left.{{ B_{\theta \rho} }}\right\vert}_{{\phi = \phi_0}}^{{\phi_1}} d\rho d\theta.\end{aligned} \hspace{\stretch{1}}(1.0.92)
There is a geometric interpretation to these oriented area integrals, especially when written out explicitly in terms of the differentials along the parameterization directions. Pulling out a sign explicitly to match the geometry (as we had to also do for the line integrals in the two parameter volume element case), we can write this as
\begin{aligned}\int_{\partial V} d^2 \mathbf{x} \cdot B = -\int {\left.{{ \left( { d\mathbf{x}_\phi \wedge d\mathbf{x}_\theta } \right) \cdot B }}\right\vert}_{{\rho = \rho_0}}^{{\rho_1}} -\int{\left.{{ \left( { d\mathbf{x}_\rho \wedge d\mathbf{x}_\phi } \right) \cdot B }}\right\vert}_{{\theta = \theta_0}}^{{\theta_1}} -\int{\left.{{ \left( { d\mathbf{x}_\theta \wedge d\mathbf{x}_\rho } \right) \cdot B }}\right\vert}_{{\phi = \phi_0}}^{{\phi_1}}.\end{aligned} \hspace{\stretch{1}}(1.0.94)
When written out in this differential form, each of the respective area elements is an oriented area along one of the faces of the parameterization volume, much like the line integral that results from a two parameter volume curl integral. This is visualized in fig. 1.8. In this figure, faces (1) and (3) are “top faces”, those with signs matching the tops of the evaluation ranges eq. 1.0.94, whereas face (2) is a bottom face with a sign that is correspondingly reversed.
Fig 1.8. Boundary faces of a spherical parameterization region
## Example: Minkowski hyperbolic-spherical polar parameterization of 3D subspace
Working with a three parameter volume element in a Minkowski space does not change much. For example in a 4D space with $\left( {\mathbf{e}_4} \right)^2 = -1$, we can employ a hyperbolic-spherical parameterization similar to that used above for the 4D Euclidean space
\begin{aligned}\mathbf{x}(x, \rho, \alpha, \phi)=\left\{ {x, \rho \sinh \alpha \cos\phi, \rho \sinh \alpha \sin\phi, \rho \cosh \alpha} \right\}=\mathbf{e}_1 x + \mathbf{e}_4 \rho \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha } \right).\end{aligned} \hspace{\stretch{1}}(1.0.95)
This has tangent space basis elements
\begin{aligned}\begin{aligned}\mathbf{x}_\rho &= \sinh\alpha \left( { \cos\phi \mathbf{e}_2 + \sin\phi \mathbf{e}_3 } \right) + \cosh\alpha \mathbf{e}_4 = \mathbf{e}_4 \exp\left( {\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha} \right) \\ \mathbf{x}_\alpha &=\rho \cosh\alpha \left( { \cos\phi \mathbf{e}_2 + \sin\phi \mathbf{e}_3} \right) + \rho \sinh\alpha \mathbf{e}_4=\rho \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \exp\left( {-\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha} \right) \\ \mathbf{x}_\phi &=\rho \sinh\alpha \left( { \mathbf{e}_3 \cos\phi - \mathbf{e}_2 \sin\phi} \right) = \rho\sinh\alpha \mathbf{e}_3 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.96)
This is a normal basis, but again not orthonormal. Specifically, for $i,j \in \left\{ {\rho, \theta, \phi} \right\}$ we have
\begin{aligned}\mathbf{x}_i \cdot \mathbf{x}_j =\begin{bmatrix}-1 & 0 & 0 \\ 0 & \rho^2 & 0 \\ 0 & 0 & \rho^2 \sinh^2 \alpha \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.97)
where we see that the radial vector $\mathbf{x}_\rho$ is timelike. We can form the dual basis again by inspection
\begin{aligned}\begin{aligned}\mathbf{x}_\rho &= -\mathbf{e}_4 \exp\left( {\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha} \right) \\ \mathbf{x}_\alpha &= \frac{1}{{\rho}} \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \exp\left( {-\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha} \right) \\ \mathbf{x}_\phi &= \frac{1}{{\rho\sinh\alpha}} \mathbf{e}_3 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.98)
The area elements are
\begin{aligned}\begin{aligned}\mathbf{x}_\alpha \wedge \mathbf{x}_\phi &=\rho^2 \sinh\alpha \left(-\mathbf{e}_4 \mathbf{e}_3 \sinh\alpha \cos\phi+\cosh\alpha \mathbf{e}_2 \mathbf{e}_3+\sinh\alpha \sin\phi \mathbf{e}_2 \mathbf{e}_4\right) \\ \mathbf{x}_\phi \wedge \mathbf{x}_\rho &=\rho \sinh\alpha \left(-\mathbf{e}_2 \mathbf{e}_3 \sinh\alpha-\mathbf{e}_2 \mathbf{e}_4 \cosh\alpha \sin\phi+\cosh\alpha \cos\phi \mathbf{e}_3 \mathbf{e}_4\right) \\ \mathbf{x}_\rho \wedge \mathbf{x}_\alpha &=-\mathbf{e}_4 \rho \left(\cos\phi \mathbf{e}_2+\sin\phi \mathbf{e}_3\right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.99)
or
\begin{aligned}\begin{aligned}\mathbf{x}_\alpha \wedge \mathbf{x}_\phi &=\rho^2 \sinh\alpha \mathbf{e}_2 \mathbf{e}_3 \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{-\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha } \right) \\ \mathbf{x}_\phi \wedge \mathbf{x}_\rho &=\rho\sinh\alpha \mathbf{e}_3 \mathbf{e}_4 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \exp\left( {\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha} \right) \\ \mathbf{x}_\rho \wedge \mathbf{x}_\alpha &=-\mathbf{e}_4 \mathbf{e}_2 \rho e^{\mathbf{e}_2 \mathbf{e}_3 \phi}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.100)
The volume element also reduces nicely, and is
\begin{aligned}\mathbf{x}_\rho \wedge \mathbf{x}_\alpha \wedge \mathbf{x}_\phi = \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4 \rho^2 \sinh\alpha.\end{aligned} \hspace{\stretch{1}}(1.0.101)
The area and volume element reductions were once again messy, done in software using \nbref{sphericalSurfaceAndVolumeElementsMinkowski.nb}. However, we really only need eq. 1.0.96 to perform the Stokes integration.
# Stokes theorem, four variable volume element parameterization
Volume elements for up to four parameters are likely of physical interest, with the four volume elements of interest for relativistic physics in $\bigwedge^{3,1}$ spaces. For example, we may wish to use a parameterization $u^1 = x, u^2 = y, u^3 = z, u^4 = \tau = c t$, with a four volume
\begin{aligned}d^4 \mathbf{x}=d\mathbf{x}_x \wedge d\mathbf{x}_y \wedge d\mathbf{x}_z \wedge d\mathbf{x}_\tau,\end{aligned} \hspace{\stretch{1}}(1.102)
We follow the same procedure to calculate the corresponding boundary surface “area” element (with dimensions of volume in this case). This is
\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= \left( { d^4 \mathbf{x} \cdot \mathbf{x}^i } \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3 du^4\left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \mathbf{x}^i } \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3 du_4\left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) {\delta_4}^i-\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_4 } \right) {\delta_3}^i+\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) {\delta_2}^i-\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) {\delta_1}^i} \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3 du^4\left( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_4 \mathbf{f}-\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_4 } \right) \cdot \partial_3 \mathbf{f}+\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \partial_2 \mathbf{f}-\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \partial_1 \mathbf{f}} \right).\end{aligned} \hspace{\stretch{1}}(1.103)
Our boundary value surface element is therefore
\begin{aligned}d^3 \mathbf{x} = \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3- \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_4+ \mathbf{x}_1 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4- \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4.\end{aligned} \hspace{\stretch{1}}(1.104)
where it is implied that this (and the dot products with $\mathbf{f}$) are evaluated on the boundaries of the integration ranges of the omitted index. This same boundary form can be used for vector, bivector and trivector variations of Stokes theorem.
# Duality and its relation to the pseudoscalar.
Looking to eq. 1.0.181 of lemma 6, and scaling the wedge product $\mathbf{a} \wedge \mathbf{b}$ by its absolute magnitude, we can express duality using that scaled bivector as a pseudoscalar for the plane that spans $\left\{ {\mathbf{a}, \mathbf{b}} \right\}$. Let’s introduce a subscript notation for such scaled blades
\begin{aligned}I_{\mathbf{a}\mathbf{b}} = \frac{\mathbf{a} \wedge \mathbf{b}}{\left\lvert {\mathbf{a} \wedge \mathbf{b}} \right\rvert}.\end{aligned} \hspace{\stretch{1}}(1.105)
This allows us to express the unit vector in the direction of $\mathbf{a}^{*}$ as
\begin{aligned}\widehat{\mathbf{a}^{*}} = \hat{\mathbf{b}} \frac{\left\lvert {\mathbf{a} \wedge \mathbf{b}} \right\rvert}{\mathbf{a} \wedge \mathbf{b}}= \hat{\mathbf{b}} \frac{1}{{I_{\mathbf{a} \mathbf{b}}}}.\end{aligned} \hspace{\stretch{1}}(1.0.106)
Following the pattern of eq. 1.0.181, it is clear how to express the dual vectors for higher dimensional subspaces. For example
or for the unit vector in the direction of $\mathbf{a}^{*}$,
\begin{aligned}\widehat{\mathbf{a}^{*}} = I_{\mathbf{b} \mathbf{c}} \frac{1}{{I_{\mathbf{a} \mathbf{b} \mathbf{c}} }}.\end{aligned}
# Divergence theorem.
When the curl integral is a scalar result we are able to apply duality relationships to obtain the divergence theorem for the corresponding space. We will be able to show that a relationship of the following form holds
\begin{aligned}\int_V dV \boldsymbol{\nabla} \cdot \mathbf{f} = \int_{\partial V} dA_i \hat{\mathbf{n}}^i \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.0.107)
Here $\mathbf{f}$ is a vector, $\hat{\mathbf{n}}^i$ is normal to the boundary surface, and $dA_i$ is the area of this bounding surface element. We wish to quantify these more precisely, especially because the orientation of the normal vectors are metric dependent. Working a few specific examples will show the pattern nicely, but it is helpful to first consider some aspects of the general case.
First note that, for a scalar Stokes integral we are integrating the vector derivative curl of a blade $F \in \bigwedge^{k-1}$ over a k-parameter volume element. Because the dimension of the space matches the number of parameters, the projection of the gradient onto the tangent space is exactly that gradient
\begin{aligned}\int_V d^k \mathbf{x} \cdot (\boldsymbol{\partial} \wedge F) =\int_V d^k \mathbf{x} \cdot (\boldsymbol{\nabla} \wedge F).\end{aligned} \hspace{\stretch{1}}(1.0.108)
Multiplication of $F$ by the pseudoscalar will always produce a vector. With the introduction of such a dual vector, as in
\begin{aligned}F = I \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.0.108)
Stokes theorem takes the form
\begin{aligned}\int_V d^k \mathbf{x} \cdot {\left\langle{{\boldsymbol{\nabla} I \mathbf{f}}}\right\rangle}_{k}= \int_{\partial V} \left\langle{{ d^{k-1} \mathbf{x} I \mathbf{f}}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(1.0.108)
or
\begin{aligned}\int_V \left\langle{{ d^k \mathbf{x} \boldsymbol{\nabla} I \mathbf{f}}}\right\rangle= \int_{\partial V} \left( { d^{k-1} \mathbf{x} I} \right) \cdot \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.0.108)
where we will see that the vector $d^{k-1} \mathbf{x} I$ can roughly be characterized as a normal to the boundary surface. Using primes to indicate the scope of the action of the gradient, cyclic permutation within the scalar selection operator can be used to factor out the pseudoscalar
\begin{aligned}\int_V \left\langle{{ d^k \mathbf{x} \boldsymbol{\nabla} I \mathbf{f}}}\right\rangle &= \int_V \left\langle{{ \mathbf{f}' d^k \mathbf{x} \boldsymbol{\nabla}' I}}\right\rangle \\ &= \int_V {\left\langle{{ \mathbf{f}' d^k \mathbf{x} \boldsymbol{\nabla}'}}\right\rangle}_{k} I \\ &= \int_V(-1)^{k+1} d^k \mathbf{x} \left( { \boldsymbol{\nabla} \cdot \mathbf{f}} \right) I \\ &= (-1)^{k+1} I^2\int_V dV\left( { \boldsymbol{\nabla} \cdot \mathbf{f}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.108)
The second last step uses lemma 8, and the last writes $d^k \mathbf{x} = I^2 \left\lvert {d^k \mathbf{x}} \right\rvert = I^2 dV$, where we have assumed (without loss of generality) that $d^k \mathbf{x}$ has the same orientation as the pseudoscalar for the space. We also assume that the parameterization is non-degenerate over the integration volume (i.e. no $d\mathbf{x}_i = 0$), so the sign of this product cannot change.
Let’s now return to the normal vector $d^{k-1} \mathbf{x} I$. With $d^{k-1} u_i = du^1 du^2 \cdots du^{i-1} du^{i+1} \cdots du^k$ (the $i$ indexed differential omitted), and $I_{ab\cdots c} = (\mathbf{x}_a \wedge \mathbf{x}_b \wedge \cdots \wedge \mathbf{x}_c)/\left\lvert {\mathbf{x}_a \wedge \mathbf{x}_b \wedge \cdots \wedge \mathbf{x}_c} \right\rvert$, we have
\begin{aligned}\begin{aligned}d^{k-1} \mathbf{x} I&=d^{k-1} u_i \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \cdots \wedge \mathbf{x}_k} \right) \cdot \mathbf{x}^i I \\ &= I_{1 2 \cdots (k-1)} I \left\lvert {d\mathbf{x}_1 \wedge d\mathbf{x}_2 \wedge \cdots \wedge d\mathbf{x}_{k-1} } \right\rvert \\ &\quad -I_{1 \cdots (k-2) k} I \left\lvert {d\mathbf{x}_1 \wedge \cdots \wedge d\mathbf{x}_{k-2} \wedge d\mathbf{x}_k} \right\rvert+ \cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.113)
We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\mathbf{a}$ with respect to the unit pseudoscalar $I_{\mathbf{b} \cdots \mathbf{c} \mathbf{d}}$ in a subspace spanned by $\left\{ {\mathbf{a}, \cdots \mathbf{c}, \mathbf{d}} \right\}$ is
\begin{aligned}\widehat{\mathbf{a}^{*}} = I_{\mathbf{b} \cdots \mathbf{c} \mathbf{d}} \frac{1}{{ I_{\mathbf{a} \cdots \mathbf{c} \mathbf{d}} }},\end{aligned} \hspace{\stretch{1}}(1.0.114)
or
\begin{aligned}\widehat{\mathbf{a}^{*}} I_{\mathbf{a} \cdots \mathbf{c} \mathbf{d}}^2=I_{\mathbf{b} \cdots \mathbf{c} \mathbf{d}}.\end{aligned} \hspace{\stretch{1}}(1.0.115)
This allows us to write
\begin{aligned}d^{k-1} \mathbf{x} I= I^2 \sum_i \widehat{\mathbf{x}^i} d{A'}_i\end{aligned} \hspace{\stretch{1}}(1.0.116)
where $d{A'}_i = \pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it.
This leaves us with
\begin{aligned}\int_V dV \boldsymbol{\nabla} \cdot \mathbf{f} = (-1)^{k+1} \int_{\partial V} d{A'}_i \widehat{\mathbf{x}^i} \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.0.117)
To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples.
## Example: 2D divergence theorem
Let’s start back at
\begin{aligned}\int_A \left\langle{{ d^2 \mathbf{x} \boldsymbol{\nabla} I \mathbf{f} }}\right\rangle = \int_{\partial A} \left( { d^1 \mathbf{x} I} \right) \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.118)
On the left our integral can be rewritten as
\begin{aligned}\int_A \left\langle{{ d^2 \mathbf{x} \boldsymbol{\nabla} I \mathbf{f} }}\right\rangle &= -\int_A \left\langle{{ d^2 \mathbf{x} I \boldsymbol{\nabla} \mathbf{f} }}\right\rangle \\ &= -\int_A d^2 \mathbf{x} I \left( { \boldsymbol{\nabla} \cdot \mathbf{f} } \right) \\ &= - I^2 \int_A dA \boldsymbol{\nabla} \cdot \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.119)
where $d^2 \mathbf{x} = I dA$ and we pick the pseudoscalar with the same orientation as the volume (area in this case) element $I = (\mathbf{x}_1 \wedge \mathbf{x}_2)/\left\lvert {\mathbf{x}_1 \wedge \mathbf{x}_2} \right\rvert$.
For the boundary form we have
\begin{aligned}d^1 \mathbf{x} = du^2 \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \mathbf{x}^1+ du^1 \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \mathbf{x}^2= -du^2 \mathbf{x}_2 +du^1 \mathbf{x}_1.\end{aligned} \hspace{\stretch{1}}(1.120)
The duality relations for the tangent space are
\begin{aligned}\begin{aligned}\mathbf{x}^2 &= \mathbf{x}_1 \frac{1}{{\mathbf{x}_2 \wedge \mathbf{x}_1}} \\ \mathbf{x}^1 &= \mathbf{x}_2 \frac{1}{{\mathbf{x}_1 \wedge \mathbf{x}_2}}\end{aligned},\end{aligned} \hspace{\stretch{1}}(1.0.121)
or
\begin{aligned}\begin{aligned}\widehat{\mathbf{x}^2} &= -\widehat{\mathbf{x}_1} \frac{1}{I} \\ \widehat{\mathbf{x}^1} &= \widehat{\mathbf{x}_2} \frac{1}{I}\end{aligned}.\end{aligned} \hspace{\stretch{1}}(1.0.122)
Back substitution into the line element gives
\begin{aligned}d^1 \mathbf{x} = -du^2 \left\lvert {\mathbf{x}_2} \right\rvert \widehat{\mathbf{x}_2}+du^1 \left\lvert {\mathbf{x}_1} \right\rvert \widehat{\mathbf{x}_1}=-du^2 \left\lvert {\mathbf{x}_2} \right\rvert \widehat{\mathbf{x}^1} I-du^1 \left\lvert {\mathbf{x}_1} \right\rvert \widehat{\mathbf{x}^2} I.\end{aligned} \hspace{\stretch{1}}(1.0.122)
Writing (no sum) $du^i \left\lvert {\mathbf{x}_i} \right\rvert = ds_i$, we have
\begin{aligned}d^1 \mathbf{x} I = -\left( { ds_2 \widehat{\mathbf{x}^1} +ds_1 \widehat{\mathbf{x}^2} } \right) I^2.\end{aligned} \hspace{\stretch{1}}(1.0.122)
This provides us a divergence and normal relationship, with $-I^2$ terms on each side that can be canceled. Restoring explicit range evaluation, that is
\begin{aligned}\int_A dA \boldsymbol{\nabla} \cdot \mathbf{f}=\int_{\Delta u^2} {\left.{{ ds_2 \widehat{\mathbf{x}^1} \cdot \mathbf{f}}}\right\vert}_{{\Delta u^1}}+ \int_{\Delta u^1} {\left.{{ ds_1 \widehat{\mathbf{x}^2} \cdot \mathbf{f}}}\right\vert}_{{\Delta u^2}}=\int_{\Delta u^2} {\left.{{ ds_2 \widehat{\mathbf{x}^1} \cdot \mathbf{f}}}\right\vert}_{{u^1(1)}}-\int_{\Delta u^2} {\left.{{ ds_2 \widehat{\mathbf{x}^1} \cdot \mathbf{f}}}\right\vert}_{{u^1(0)}}+ \int_{\Delta u^1} {\left.{{ ds_1 \widehat{\mathbf{x}^2} \cdot \mathbf{f}}}\right\vert}_{{u^2(0)}}- \int_{\Delta u^1} {\left.{{ ds_1 \widehat{\mathbf{x}^2} \cdot \mathbf{f}}}\right\vert}_{{u^2(0)}}.\end{aligned} \hspace{\stretch{1}}(1.0.122)
Let’s consider this graphically for an Euclidean metric as illustrated in fig. 1.9.
Fig 1.9. Normals on area element
We see that
1. along $u^2(0)$ the outwards normal is $-\widehat{\mathbf{x}^2}$,
2. along $u^2(1)$ the outwards normal is $\widehat{\mathbf{x}^2}$,
3. along $u^1(0)$ the outwards normal is $-\widehat{\mathbf{x}^1}$, and
4. along $u^1(1)$ the outwards normal is $\widehat{\mathbf{x}^2}$.
Writing that outwards normal as $\hat{\mathbf{n}}$, we have
\begin{aligned}\int_A dA \boldsymbol{\nabla} \cdot \mathbf{f}= \mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} ds \hat{\mathbf{n}} \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.0.126)
Note that we can use the same algebraic notion of outward normal for non-Euclidean spaces, although cannot expect the geometry to look anything like that of the figure.
## Example: 3D divergence theorem
As with the 2D example, let’s start back with
\begin{aligned}\int_V \left\langle{{ d^3 \mathbf{x} \boldsymbol{\nabla} I \mathbf{f} }}\right\rangle = \int_{\partial V} \left( { d^2 \mathbf{x} I} \right) \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.127)
In a 3D space, the pseudoscalar commutes with all grades, so we have
\begin{aligned}\int_V \left\langle{{ d^3 \mathbf{x} \boldsymbol{\nabla} I \mathbf{f} }}\right\rangle=\int_V \left( { d^3 \mathbf{x} I } \right) \boldsymbol{\nabla} \cdot \mathbf{f}=I^2 \int_V dV \boldsymbol{\nabla} \cdot \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.128)
where $d^3 \mathbf{x} I = dV I^2$, and we have used a pseudoscalar with the same orientation as the volume element
\begin{aligned}\begin{aligned}I &= \widehat{ \mathbf{x}_{123} } \\ \mathbf{x}_{123} &= \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.129)
In the boundary integral our dual two form is
\begin{aligned}d^2 \mathbf{x} I= du^1 du^2 \mathbf{x}_1 \wedge \mathbf{x}_2+du^3 du^1 \mathbf{x}_3 \wedge \mathbf{x}_1+du^2 du^3 \mathbf{x}_2 \wedge \mathbf{x}_3= \left( { dA_{3} \widehat{ \mathbf{x}_{12} } \frac{1}{I}+dA_{2} \widehat{ \mathbf{x}_{31} } \frac{1}{I}+dA_{1} \widehat{ \mathbf{x}_{23} } \frac{1}{I}} \right) I^2,\end{aligned} \hspace{\stretch{1}}(1.0.129)
where $\mathbf{x}_{ij} = \mathbf{x}_i \wedge \mathbf{x}_j$, and
\begin{aligned}\begin{aligned}dA_1 &= \left\lvert {d\mathbf{x}_2 \wedge d\mathbf{x}_3} \right\rvert \\ dA_2 &= \left\lvert {d\mathbf{x}_3 \wedge d\mathbf{x}_1} \right\rvert \\ dA_3 &= \left\lvert {d\mathbf{x}_1 \wedge d\mathbf{x}_2} \right\rvert.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.131)
Observe that we can do a cyclic permutation of a 3 blade without any change of sign, for example
\begin{aligned}\mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 =-\mathbf{x}_2 \wedge \mathbf{x}_1 \wedge \mathbf{x}_3 =\mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_1.\end{aligned} \hspace{\stretch{1}}(1.0.132)
Because of this we can write the dual two form as we expressed the normals in lemma 7
\begin{aligned}d^2 \mathbf{x} I = \left( { dA_1 \widehat{\mathbf{x}_{23}} \frac{1}{{\widehat{\mathbf{x}_{123}}}} + dA_2 \widehat{\mathbf{x}_{31}} \frac{1}{{\widehat{\mathbf{x}_{231}}}} + dA_3 \widehat{\mathbf{x}_{12}} \frac{1}{{\widehat{\mathbf{x}_{312}}}}} \right) I^2=\left( { dA_1 \widehat{\mathbf{x}^1}+dA_2 \widehat{\mathbf{x}^2}+dA_3 \widehat{\mathbf{x}^3} } \right) I^2.\end{aligned} \hspace{\stretch{1}}(1.0.132)
We can now state the 3D divergence theorem, canceling out the metric and orientation dependent term $I^2$ on both sides
\begin{aligned}\int_V dV \boldsymbol{\nabla} \cdot \mathbf{f}=\int dA \hat{\mathbf{n}} \cdot \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.0.134)
where (sums implied)
\begin{aligned}dA \hat{\mathbf{n}} = dA_i \widehat{\mathbf{x}^i},\end{aligned} \hspace{\stretch{1}}(1.0.135)
and
\begin{aligned}\begin{aligned}{\left.{{\hat{\mathbf{n}}}}\right\vert}_{{u^i = u^i(1)}} &= \widehat{\mathbf{x}^i} \\ {\left.{{\hat{\mathbf{n}}}}\right\vert}_{{u^i = u^i(0)}} &= -\widehat{\mathbf{x}^i}\end{aligned}.\end{aligned} \hspace{\stretch{1}}(1.0.136)
The outwards normals at the upper integration ranges of a three parameter surface are depicted in fig. 1.10.
Fig 1.10. Outwards normals on volume at upper integration ranges.
This sign alternation originates with the two form elements $\left( {d\mathbf{x}_i \wedge d\mathbf{x}_j} \right) \cdot F$ from the Stokes boundary integral, which were explicitly evaluated at the endpoints of the integral. That is, for $k \ne i,j$,
\begin{aligned}\int_{\partial V} \left( { d\mathbf{x}_i \wedge d\mathbf{x}_j } \right) \cdot F\equiv\int_{\Delta u^i} \int_{\Delta u^j} {\left.{{\left( { \left( { d\mathbf{x}_i \wedge d\mathbf{x}_j } \right) \cdot F } \right)}}\right\vert}_{{u^k = u^k(1)}}-{\left.{{\left( { \left( { d\mathbf{x}_i \wedge d\mathbf{x}_j } \right) \cdot F } \right)}}\right\vert}_{{u^k = u^k(0)}}\end{aligned} \hspace{\stretch{1}}(1.0.137)
In the context of the divergence theorem, this means that we are implicitly requiring the dot products $\widehat{\mathbf{x}^k} \cdot \mathbf{f}$ to be evaluated specifically at the end points of the integration where $u^k = u^k(1), u^k = u^k(0)$, accounting for the alternation of sign required to describe the normals as uniformly outwards.
## Example: 4D divergence theorem
Applying Stokes theorem to a trivector $T = I \mathbf{f}$ in the 4D case we find
\begin{aligned}-I^2 \int_V d^4 x \boldsymbol{\nabla} \cdot \mathbf{f} = \int_{\partial V} \left( { d^3 \mathbf{x} I} \right) \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.138)
Here the pseudoscalar has been picked to have the same orientation as the hypervolume element $d^4 \mathbf{x} = I d^4 x$. Writing $\mathbf{x}_{ij \cdots k} = \mathbf{x}_i \wedge \mathbf{x}_j \wedge \cdots \mathbf{x}_k$ the dual of the three form is
\begin{aligned}d^3 \mathbf{x} I &= \left( { du^1 du^2 du^3 \mathbf{x}_{123}-du^1 du^2 du^4 \mathbf{x}_{124}+du^1 du^3 du^4 \mathbf{x}_{134}-du^2 du^3 du^4 \mathbf{x}_{234}} \right) I \\ &= \left( { dA^{123} \widehat{ \mathbf{x}_{123} } -dA^{124} \widehat{ \mathbf{x}_{124} } +dA^{134} \widehat{ \mathbf{x}_{134} } -dA^{234} \widehat{ \mathbf{x}_{234} }} \right) I \\ &= \left( { dA^{123} \widehat{ \mathbf{x}_{123} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}} -dA^{124} \widehat{ \mathbf{x}_{124} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}} +dA^{134} \widehat{ \mathbf{x}_{134} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}} -dA^{234} \widehat{ \mathbf{x}_{234} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}}} \right) I^2 \\ &= -\left( { dA^{123} \widehat{ \mathbf{x}_{123} } \frac{1}{{\widehat{\mathbf{x}_{4123} }}} +dA^{124} \widehat{ \mathbf{x}_{124} } \frac{1}{{\widehat{\mathbf{x}_{3412} }}} +dA^{134} \widehat{ \mathbf{x}_{134} } \frac{1}{{\widehat{\mathbf{x}_{2341} }}} +dA^{234} \widehat{ \mathbf{x}_{234} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}}} \right) I^2 \\ &= -\left( { dA^{123} \widehat{ \mathbf{x}_{123} } \frac{1}{{\widehat{\mathbf{x}_{4123} }}} +dA^{124} \widehat{ \mathbf{x}_{412} } \frac{1}{{\widehat{\mathbf{x}_{3412} }}} +dA^{134} \widehat{ \mathbf{x}_{341} } \frac{1}{{\widehat{\mathbf{x}_{2341} }}} +dA^{234} \widehat{ \mathbf{x}_{234} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}}} \right) I^2 \\ &= -\left( { dA^{123} \widehat{ \mathbf{x}^{4} } +dA^{124} \widehat{ \mathbf{x}^{3} } +dA^{134} \widehat{ \mathbf{x}^{2} } +dA^{234} \widehat{ \mathbf{x}^{1} } } \right) I^2\end{aligned} \hspace{\stretch{1}}(1.139)
Here, we’ve written
\begin{aligned}dA^{ijk} = \left\lvert { d\mathbf{x}_i \wedge d\mathbf{x}_j \wedge d\mathbf{x}_k } \right\rvert.\end{aligned} \hspace{\stretch{1}}(1.140)
Observe that the dual representation nicely removes the alternation of sign that we had in the Stokes theorem boundary integral, since each alternation of the wedged vectors in the pseudoscalar changes the sign once.
As before, we define the outwards normals as $\hat{\mathbf{n}} = \pm \widehat{\mathbf{x}^i}$ on the upper and lower integration ranges respectively. The scalar area elements on these faces can be written in a dual form
\begin{aligned}\begin{aligned} dA_4 &= dA^{123} \\ dA_3 &= dA^{124} \\ dA_2 &= dA^{134} \\ dA_1 &= dA^{234} \end{aligned},\end{aligned} \hspace{\stretch{1}}(1.0.141)
so that the 4D divergence theorem looks just like the 2D and 3D cases
\begin{aligned}\int_V d^4 x \boldsymbol{\nabla} \cdot \mathbf{f} = \int_{\partial V} d^3 x \hat{\mathbf{n}} \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.0.142)
Here we define the volume scaled normal as
\begin{aligned}d^3 x \hat{\mathbf{n}} = dA_i \widehat{\mathbf{x}^i}.\end{aligned} \hspace{\stretch{1}}(1.0.143)
As before, we have made use of the implicit fact that the three form (and it’s dot product with $\mathbf{f}$) was evaluated on the boundaries of the integration region, with a toggling of sign on the lower limit of that evaluation that is now reflected in what we have defined as the outwards normal.
We also obtain explicit instructions from this formalism how to compute the “outwards” normal for this surface in a 4D space (unit scaling of the dual basis elements), something that we cannot compute using any sort of geometrical intuition. For free we’ve obtained a result that applies to both Euclidean and Minkowski (or other non-Euclidean) spaces.
# Volume integral coordinate representations
It may be useful to formulate the curl integrals in tensor form. For vectors $\mathbf{f}$, and bivectors $B$, the coordinate representations of those differential forms (\cref{pr:stokesTheoremGeometricAlgebraII:1}) are
\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=- d^2 u \epsilon^{ a b } \partial_a f_b\end{aligned} \hspace{\stretch{1}}(1.0.144a)
\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=-d^3 u \epsilon^{a b c} \mathbf{x}_a \partial_b f_{c}\end{aligned} \hspace{\stretch{1}}(1.0.144b)
\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=-\frac{1}{2} d^4 u \epsilon^{a b c d} \mathbf{x}_a \wedge \mathbf{x}_b \partial_{c} f_{d}\end{aligned} \hspace{\stretch{1}}(1.0.144c)
\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)=-\frac{1}{2}d^3 u \epsilon^{a b c} \partial_a B_{b c}\end{aligned} \hspace{\stretch{1}}(1.0.144d)
\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)=-\frac{1}{2} d^4 u \epsilon^{a b c d} \mathbf{x}_a \partial_b B_{cd}\end{aligned} \hspace{\stretch{1}}(1.0.144e)
\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge T } \right)=-d^4 u\left( {\partial_4 T_{123}-\partial_3 T_{124}+\partial_2 T_{134}-\partial_1 T_{234}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.144f)
Here the bivector $B$ and trivector $T$ is expressed in terms of their curvilinear components on the tangent space
\begin{aligned}B = \frac{1}{2} \mathbf{x}^i \wedge \mathbf{x}^j B_{ij} + B_\perp\end{aligned} \hspace{\stretch{1}}(1.0.145a)
\begin{aligned}T = \frac{1}{{3!}} \mathbf{x}^i \wedge \mathbf{x}^j \wedge \mathbf{x}^k T_{ijk} + T_\perp,\end{aligned} \hspace{\stretch{1}}(1.0.145b)
where
\begin{aligned}B_{ij} = \mathbf{x}_j \cdot \left( { \mathbf{x}_i \cdot B } \right) = -B_{ji}.\end{aligned} \hspace{\stretch{1}}(1.0.146a)
\begin{aligned}T_{ijk} = \mathbf{x}_k \cdot \left( { \mathbf{x}_j \cdot \left( { \mathbf{x}_i \cdot B } \right)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.146b)
For the trivector components are also antisymmetric, changing sign with any interchange of indices.
Note that eq. 1.0.144d and eq. 1.0.144f appear much different on the surface, but both have the same structure. This can be seen by writing for former as
\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)=-d^3 u\left( { \partial_1 B_{2 3} + \partial_2 B_{3 1} + \partial_3 B_{1 2}} \right)=-d^3 u\left( { \partial_3 B_{1 2} - \partial_2 B_{1 3} + \partial_1 B_{2 3}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.146b)
In both of these we have an alternation of sign, where the tensor index skips one of the volume element indices is sequence. We’ve seen in the 4D divergence theorem that this alternation of sign can be related to a duality transformation.
In integral form (no sum over indexes $i$ in $du^i$ terms), these are
\begin{aligned}\int d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=- \epsilon^{ a b } \int {\left.{{du^b f_b}}\right\vert}_{{\Delta u^a}}\end{aligned} \hspace{\stretch{1}}(1.0.148a)
\begin{aligned}\int d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=-\epsilon^{a b c} \int du^a du^c{\left.{{\mathbf{x}_a f_{c}}}\right\vert}_{{\Delta u^b}}\end{aligned} \hspace{\stretch{1}}(1.0.148b)
\begin{aligned}\int d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=-\frac{1}{2} \epsilon^{a b c d} \int du^a du^b du^d{\left.{{\mathbf{x}_a \wedge \mathbf{x}_b f_{d}}}\right\vert}_{{\Delta u^c}}\end{aligned} \hspace{\stretch{1}}(1.0.148c)
\begin{aligned}\int d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)=-\frac{1}{2}\epsilon^{a b c} \int du^b du^c{\left.{{B_{b c}}}\right\vert}_{{\Delta u^a}}\end{aligned} \hspace{\stretch{1}}(1.0.148d)
\begin{aligned}\int d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)=-\frac{1}{2} \epsilon^{a b c d} \int du^a du^c du^d{\left.{{\mathbf{x}_a B_{cd}}}\right\vert}_{{\Delta u^b}}\end{aligned} \hspace{\stretch{1}}(1.0.148e)
\begin{aligned}\int d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge T } \right)=-\int \left( {du^1 du^2 du^3 {\left.{{T_{123}}}\right\vert}_{{\Delta u^4}}-du^1 du^2 du^4 {\left.{{T_{124}}}\right\vert}_{{\Delta u^3}}+du^1 du^3 du^4 {\left.{{T_{134}}}\right\vert}_{{\Delta u^2}}-du^2 du^3 du^4 {\left.{{T_{234}}}\right\vert}_{{\Delta u^1}}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.148f)
Of these, I suspect that only eq. 1.0.148a and eq. 1.0.148d are of use.
# Final remarks
Because we have used curvilinear coordinates from the get go, we have arrived naturally at a formulation that works for both Euclidean and non-Euclidean geometries, and have demonstrated that Stokes (and the divergence theorem) holds regardless of the geometry or the parameterization. We also know explicitly how to formulate both theorems for any parameterization that we choose, something much more valuable than knowledge that this is possible.
For the divergence theorem we have introduced the concept of outwards normal (for example in 3D, eq. 1.0.136), which still holds for non-Euclidean geometries. We may not be able to form intuitive geometrical interpretations for these normals, but do have an algebraic description of them.
# Appendix
## Question: Expand volume elements in coordinates
Show that the coordinate representation for the volume element dotted with the curl can be represented as a sum of antisymmetric terms. That is
• (a)Prove eq. 1.0.144a
• (b)Prove eq. 1.0.144b
• (c)Prove eq. 1.0.144c
• (d)Prove eq. 1.0.144d
• (e)Prove eq. 1.0.144e
• (f)Prove eq. 1.0.144f
### (a) Two parameter volume, curl of vector
\begin{aligned}d^2 \mathbf{x} \cdot \left( \boldsymbol{\partial} \wedge \mathbf{f} \right) &= d^2 u\Bigl( { \left( \mathbf{x}_1 \wedge \mathbf{x}_2 \right) \cdot \mathbf{x}^i } \Bigr) \cdot \partial_i \mathbf{f} \\ &= d^2 u \left( \mathbf{x}_1 \cdot \partial_2 \mathbf{f}-\mathbf{x}_2 \cdot \partial_1 \mathbf{f} \right) \\ &= d^2 u\left( \partial_2 f_1-\partial_1 f_2 \right) \\ &= - d^2 u \epsilon^{ab} \partial_{a} f_{b}. \qquad\square\end{aligned} \hspace{\stretch{1}}(1.149)
### (b) Three parameter volume, curl of vector
\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= d^3 u\Bigl( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \mathbf{x}^i } \Bigr) \cdot \partial_i \mathbf{f} \\ &= d^3 u\Bigl( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \partial_3 \mathbf{f}+\left( { \mathbf{x}_3 \wedge \mathbf{x}_1 } \right) \cdot \partial_2 \mathbf{f}+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_1 \mathbf{f}} \Bigr) \\ &= d^3 u\Bigl( {\left( { \mathbf{x}_1 \partial_3 \mathbf{f} \cdot \mathbf{x}_2 -\mathbf{x}_2 \partial_3 \mathbf{f} \cdot \mathbf{x}_1 } \right)+\left( { \mathbf{x}_3 \partial_2 \mathbf{f} \cdot \mathbf{x}_1 -\mathbf{x}_1 \partial_2 \mathbf{f} \cdot \mathbf{x}_3 } \right)+\left( { \mathbf{x}_2 \partial_1 \mathbf{f} \cdot \mathbf{x}_3 -\mathbf{x}_3 \partial_1 \mathbf{f} \cdot \mathbf{x}_2 } \right)} \Bigr) \\ &= d^3 u\Bigl( {\mathbf{x}_1 \left( { -\partial_2 \mathbf{f} \cdot \mathbf{x}_3 + \partial_3 \mathbf{f} \cdot \mathbf{x}_2 } \right)+\mathbf{x}_2 \left( { -\partial_3 \mathbf{f} \cdot \mathbf{x}_1 + \partial_1 \mathbf{f} \cdot \mathbf{x}_3 } \right)+\mathbf{x}_3 \left( { -\partial_1 \mathbf{f} \cdot \mathbf{x}_2 + \partial_2 \mathbf{f} \cdot \mathbf{x}_1 } \right)} \Bigr) \\ &= d^3 u\Bigl( {\mathbf{x}_1 \left( { -\partial_2 f_3 + \partial_3 f_2 } \right)+\mathbf{x}_2 \left( { -\partial_3 f_1 + \partial_1 f_3 } \right)+\mathbf{x}_3 \left( { -\partial_1 f_2 + \partial_2 f_1 } \right)} \Bigr) \\ &= - d^3 u \epsilon^{abc} \partial_b f_c. \qquad\square\end{aligned} \hspace{\stretch{1}}(1.150)
### (c) Four parameter volume, curl of vector
\begin{aligned}\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)&=d^4 u\Bigl( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \mathbf{x}^i } \Bigr) \cdot \partial_i \mathbf{f} \\ &=d^4 u\Bigl( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_4 \mathbf{f}-\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_4 } \right) \cdot \partial_3 \mathbf{f}+\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \partial_2 \mathbf{f}-\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \partial_1 \mathbf{f}} \Bigr) \\ &=d^4 u\Bigl( { \\ &\quad\quad \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \mathbf{x}_3 \cdot \partial_4 \mathbf{f}-\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 } \right) \mathbf{x}_2 \cdot \partial_4 \mathbf{f}+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \mathbf{x}_1 \cdot \partial_4 \mathbf{f} \\ &\quad-\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \mathbf{x}_4 \cdot \partial_3 \mathbf{f}+\left( { \mathbf{x}_1 \wedge \mathbf{x}_4 } \right) \mathbf{x}_2 \cdot \partial_3 \mathbf{f}-\left( { \mathbf{x}_2 \wedge \mathbf{x}_4 } \right) \mathbf{x}_1 \cdot \partial_3 \mathbf{f} \\ &\quad+ \left( { \mathbf{x}_1 \wedge \mathbf{x}_3 } \right) \mathbf{x}_4 \cdot \partial_2 \mathbf{f}-\left( { \mathbf{x}_1 \wedge \mathbf{x}_4 } \right) \mathbf{x}_3 \cdot \partial_2 \mathbf{f}+\left( { \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \mathbf{x}_1 \cdot \partial_2 \mathbf{f} \\ &\quad-\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \mathbf{x}_4 \cdot \partial_1 \mathbf{f}+\left( { \mathbf{x}_2 \wedge \mathbf{x}_4 } \right) \mathbf{x}_3 \cdot \partial_1 \mathbf{f}-\left( { \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \mathbf{x}_2 \cdot \partial_1 \mathbf{f} \\ &\qquad} \Bigr) \\ &=d^4 u\Bigl( {\mathbf{x}_1 \wedge \mathbf{x}_2 \partial_{[4} f_{3]}+\mathbf{x}_1 \wedge \mathbf{x}_3 \partial_{[2} f_{4]}+\mathbf{x}_1 \wedge \mathbf{x}_4 \partial_{[3} f_{2]}+\mathbf{x}_2 \wedge \mathbf{x}_3 \partial_{[4} f_{1]}+\mathbf{x}_2 \wedge \mathbf{x}_4 \partial_{[1} f_{3]}+\mathbf{x}_3 \wedge \mathbf{x}_4 \partial_{[2} f_{1]}} \Bigr) \\ &=- \frac{1}{2} d^4 u \epsilon^{abcd} \mathbf{x}_a \wedge \mathbf{x}_b \partial_{c} f_{d}. \qquad\square\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.151)
### (d) Three parameter volume, curl of bivector
\begin{aligned}\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)&=d^3 u\Bigl( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \mathbf{x}^i } \Bigr) \cdot \partial_i B \\ &=d^3 u\Bigl( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \partial_3 B+\left( { \mathbf{x}_3 \wedge \mathbf{x}_1 } \right) \cdot \partial_2 B+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_1 B} \Bigr) \\ &=\frac{1}{2} d^3 u\Bigl( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \partial_3 B } \right) -\mathbf{x}_2 \cdot \left( { \mathbf{x}_1 \cdot \partial_3 B } \right) \\ &\qquad +\mathbf{x}_3 \cdot \left( { \mathbf{x}_1 \cdot \partial_2 B } \right) -\mathbf{x}_1 \cdot \left( { \mathbf{x}_3 \cdot \partial_2 B } \right) \\ &\qquad +\mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot \partial_1 B } \right) -\mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot \partial_1 B } \right)} \Bigr) \\ &=\frac{1}{2} d^3 u\Bigl( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \partial_3 B - \mathbf{x}_3 \cdot \partial_2 B } \right) \\ &\qquad +\mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot \partial_1 B - \mathbf{x}_1 \cdot \partial_3 B } \right) \\ &\qquad +\mathbf{x}_3 \cdot \left( { \mathbf{x}_1 \cdot \partial_2 B - \mathbf{x}_2 \cdot \partial_1 B } \right)} \Bigr) \\ &=\frac{1}{2} d^3 u\Bigl( {\mathbf{x}_1 \cdot \left( { \partial_3 \left( { \mathbf{x}_2 \cdot B} \right) - \partial_2 \left( { \mathbf{x}_3 \cdot B} \right) } \right) \\ &\qquad +\mathbf{x}_2 \cdot \left( { \partial_1 \left( { \mathbf{x}_3 \cdot B} \right) - \partial_3 \left( { \mathbf{x}_1 \cdot B} \right) } \right) \\ &\qquad +\mathbf{x}_3 \cdot \left( { \partial_2 \left( { \mathbf{x}_1 \cdot B} \right) - \partial_1 \left( { \mathbf{x}_2 \cdot B} \right) } \right)} \Bigr) \\ &=\frac{1}{2} d^3 u\Bigl( {\partial_2 \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_1 \cdot B} \right) } \right) - \partial_3 \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_1 \cdot B} \right) } \right) \\ &\qquad+ \partial_3 \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot B} \right) } \right) - \partial_1 \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot B} \right) } \right) \\ &\qquad+ \partial_1 \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot B} \right) } \right) - \partial_2 \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_3 \cdot B} \right) } \right)} \Bigr) \\ &=\frac{1}{2} d^3 u\Bigl( {\partial_2 B_{13} - \partial_3 B_{12}+\partial_3 B_{21} - \partial_1 B_{23}+\partial_1 B_{32} - \partial_2 B_{31}} \Bigr) \\ &=d^3 u\Bigl( {\partial_2 B_{13}+\partial_3 B_{21}+\partial_1 B_{32}} \Bigr) \\ &= - \frac{1}{2} d^3 u \epsilon^{abc} \partial_a B_{bc}. \qquad\square\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.152)
### (e) Four parameter volume, curl of bivector
To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\mathbf{x}_{abc}$ for $\mathbf{x}_a \wedge \mathbf{x}_b \wedge \mathbf{x}_c$. The expansion is
\begin{aligned}\begin{aligned}d^4 \mathbf{x} \cdot \left( \boldsymbol{\partial} \wedge B \right) &= d^4 u \left( \mathbf{x}_{1234} \cdot \mathbf{x}^i \right) \cdot \partial_i B \\ &= d^4 u\left( \mathbf{x}_{123} \cdot \partial_4 B - \mathbf{x}_{124} \cdot \partial_3 B + \mathbf{x}_{134} \cdot \partial_2 B - \mathbf{x}_{234} \cdot \partial_1 B \right) \\ &= d^4 u \Bigl( \mathbf{x}_1 \left( \mathbf{x}_{23} \cdot \partial_4 B \right) + \mathbf{x}_2 \left( \mathbf{x}_{32} \cdot \partial_4 B \right) + \mathbf{x}_3 \left( \mathbf{x}_{12} \cdot \partial_4 B \right) \\ &\qquad - \mathbf{x}_1 \left( \mathbf{x}_{24} \cdot \partial_3 B \right) - \mathbf{x}_2 \left( \mathbf{x}_{41} \cdot \partial_3 B \right) - \mathbf{x}_4 \left( \mathbf{x}_{12} \cdot \partial_3 B \right) \\ &\qquad + \mathbf{x}_1 \left( \mathbf{x}_{34} \cdot \partial_2 B \right) + \mathbf{x}_3 \left( \mathbf{x}_{41} \cdot \partial_2 B \right) + \mathbf{x}_4 \left( \mathbf{x}_{13} \cdot \partial_2 B \right) \\ &\qquad - \mathbf{x}_2 \left( \mathbf{x}_{34} \cdot \partial_1 B \right) - \mathbf{x}_3 \left( \mathbf{x}_{42} \cdot \partial_1 B \right) - \mathbf{x}_4 \left( \mathbf{x}_{23} \cdot \partial_1 B \right)} \Bigr) \\ &= d^4 u \Bigl( \mathbf{x}_1 \left( \mathbf{x}_{23} \cdot \partial_4 B + \mathbf{x}_{42} \cdot \partial_3 B + \mathbf{x}_{34} \cdot \partial_2 B \right) \\ &\qquad + \mathbf{x}_2 \left( \mathbf{x}_{32} \cdot \partial_4 B + \mathbf{x}_{14} \cdot \partial_3 B + \mathbf{x}_{43} \cdot \partial_1 B \right) \\ &\qquad + \mathbf{x}_3 \left( \mathbf{x}_{12} \cdot \partial_4 B + \mathbf{x}_{41} \cdot \partial_2 B + \mathbf{x}_{24} \cdot \partial_1 B \right) \\ &\qquad + \mathbf{x}_4 \left( \mathbf{x}_{21} \cdot \partial_3 B + \mathbf{x}_{13} \cdot \partial_2 B + \mathbf{x}_{32} \cdot \partial_1 B \right)} \Bigr) \\ &= - \frac{1}{2} d^4 u \epsilon^{a b c d} \mathbf{x}_a \partial_b B_{c d}. \qquad\square\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.153)
This last step uses an intermediate result from the eq. 1.0.152 expansion above, since each of the four terms has the same structure we have previously observed.
### (f) Four parameter volume, curl of trivector
Using the $\mathbf{x}_{ijk}$ shorthand again, the initial expansion gives
\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge T } \right)=d^4 u\left( {\mathbf{x}_{123} \cdot \partial_4 T - \mathbf{x}_{124} \cdot \partial_3 T + \mathbf{x}_{134} \cdot \partial_2 T - \mathbf{x}_{234} \cdot \partial_1 T} \right).\end{aligned} \hspace{\stretch{1}}(1.0.153)
Applying lemma 4 to expand the inner products within the braces we have
\begin{aligned}\begin{aligned}\mathbf{x}_{123} \cdot \partial_4 T-&\mathbf{x}_{124} \cdot \partial_3 T+\mathbf{x}_{134} \cdot \partial_2 T-\mathbf{x}_{234} \cdot \partial_1 T \\ &=\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot \partial_4 T } \right) } \right)-\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_4 \cdot \partial_3 T } \right) } \right) \\ &\quad +\underbrace{\mathbf{x}_1 \cdot \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \partial_2 T } \right) } \right)-\mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \partial_1 T } \right) } \right)}_{\text{Apply cyclic permutations}}\\ &=\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot \partial_4 T } \right) } \right)-\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_4 \cdot \partial_3 T } \right) } \right) \\ &\quad +\mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_1 \cdot \partial_2 T } \right) } \right)-\mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_2 \cdot \partial_1 T } \right) } \right) \\ &=\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot\left( {\mathbf{x}_3 \cdot \partial_4 T-\mathbf{x}_4 \cdot \partial_3 T} \right) } \right) \\ &\quad +\mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \left( {\mathbf{x}_1 \cdot \partial_2 T-\mathbf{x}_2 \cdot \partial_1 T} \right) } \right) \\ &=\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot\left( {\partial_4 \left( { \mathbf{x}_3 \cdot T } \right)-\partial_3 \left( { \mathbf{x}_4 \cdot T } \right)} \right) } \right) \\ &\quad +\mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \left( {\partial_2 \left( { \mathbf{x}_1 \cdot T } \right)-\partial_1 \left( { \mathbf{x}_2 \cdot T } \right)} \right) } \right) \\ &=\mathbf{x}_1 \cdot \partial_4 \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot T } \right) } \right)+\mathbf{x}_2 \cdot \partial_3 \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_4 \cdot T } \right) } \right) \\ &\quad +\mathbf{x}_3 \cdot \partial_2 \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_1 \cdot T } \right) } \right)+\mathbf{x}_4 \cdot \partial_1 \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot T } \right) } \right) \\ &-\mathbf{x}_1 \cdot \left( { \left( { \partial_4 \mathbf{x}_2} \right) \cdot \left( { \mathbf{x}_3 \cdot T } \right) } \right)-\mathbf{x}_2 \cdot \left( { \left( { \partial_3 \mathbf{x}_1} \right) \cdot \left( { \mathbf{x}_4 \cdot T } \right) } \right) \\ &\quad -\mathbf{x}_3 \cdot \left( { \left( { \partial_2 \mathbf{x}_4} \right) \cdot \left( { \mathbf{x}_1 \cdot T } \right) } \right)-\mathbf{x}_4 \cdot \left( { \left( { \partial_1 \mathbf{x}_3} \right) \cdot \left( { \mathbf{x}_2 \cdot T } \right) } \right) \\ &=\mathbf{x}_1 \cdot \partial_4 \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot T } \right) } \right)+\mathbf{x}_2 \cdot \partial_3 \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_4 \cdot T } \right) } \right) \\ &\quad +\mathbf{x}_3 \cdot \partial_2 \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_1 \cdot T } \right) } \right)+\mathbf{x}_4 \cdot \partial_1 \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot T } \right) } \right) \\ &+\frac{\partial^2 \mathbf{x}}{\partial u^4 \partial u^2}\cdot\not{{\left( {\mathbf{x}_1 \cdot \left( { \mathbf{x}_3 \cdot T } \right)+\mathbf{x}_3 \cdot \left( { \mathbf{x}_1 \cdot T } \right)} \right)}} \\ &\quad +\frac{\partial^2 \mathbf{x}}{\partial u^1 \partial u^3}\cdot\not{{\left( {\mathbf{x}_2 \cdot \left( { \mathbf{x}_4 \cdot T } \right)+\mathbf{x}_4 \cdot \left( { \mathbf{x}_2 \cdot T } \right)} \right)}}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.155)
We can cancel those last terms using lemma 5. Using the same reverse chain rule expansion once more we have
\begin{aligned}\begin{aligned}\mathbf{x}_{123} \cdot \partial_4 T-&\mathbf{x}_{124} \cdot \partial_3 T+\mathbf{x}_{134} \cdot \partial_2 T-\mathbf{x}_{234} \cdot \partial_1 T \\ &=\partial_4 \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot T } \right) } \right) } \right)+\partial_3 \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_4 \cdot T } \right) } \right) } \right)+\partial_2 \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_1 \cdot T } \right) } \right) } \right)+\partial_1 \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot T } \right) } \right) } \right) \\ &-\left( { \partial_4 \mathbf{x}_1} \right)\cdot\not{{\left( {\mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot T } \right)+\mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot T } \right)} \right)}}-\left( { \partial_3 \mathbf{x}_2} \right) \cdot\not{{\left( {\mathbf{x}_1 \cdot \left( { \mathbf{x}_4 \cdot T } \right)\mathbf{x}_4 \cdot \left( { \mathbf{x}_1 \cdot T } \right)} \right)}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.156)
or
\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge T } \right)=d^4 u\Bigl( {\partial_4 T_{3 2 1}+\partial_3 T_{4 1 2}+\partial_2 T_{1 4 3}+\partial_1 T_{2 3 4}} \Bigr).\end{aligned} \hspace{\stretch{1}}(1.0.156)
The final result follows after permuting the indices slightly.
### Lemma 1. Distribution of inner products
Given two blades $A_s, B_r$ with grades subject to $s > r > 0$, and a vector $b$, the inner product distributes according to
\begin{aligned}A_s \cdot \left( { b \wedge B_r } \right) = \left( { A_s \cdot b } \right) \cdot B_r.\end{aligned}
This will allow us, for example, to expand a general inner product of the form $d^k \mathbf{x} \cdot (\boldsymbol{\partial} \wedge F)$.
The proof is straightforward, but also mechanical. Start by expanding the wedge and dot products within a grade selection operator
\begin{aligned}A_s \cdot \left( { b \wedge B_r } \right)={\left\langle{{A_s (b \wedge B_r)}}\right\rangle}_{{s - (r + 1)}}=\frac{1}{2} {\left\langle{{A_s \left( {b B_r + (-1)^{r} B_r b} \right) }}\right\rangle}_{{s - (r + 1)}}\end{aligned} \hspace{\stretch{1}}(1.158)
Solving for $B_r b$ in
\begin{aligned}2 b \cdot B_r = b B_r - (-1)^{r} B_r b,\end{aligned} \hspace{\stretch{1}}(1.159)
we have
\begin{aligned}A_s \cdot \left( { b \wedge B_r } \right)=\frac{1}{2} {\left\langle{{ A_s b B_r + A_s \left( { b B_r - 2 b \cdot B_r } \right) }}\right\rangle}_{{s - (r + 1)}}={\left\langle{{ A_s b B_r }}\right\rangle}_{{s - (r + 1)}}-\not{{{\left\langle{{ A_s \left( { b \cdot B_r } \right) }}\right\rangle}_{{s - (r + 1)}}}}.\end{aligned} \hspace{\stretch{1}}(1.160)
The last term above is zero since we are selecting the $s - r - 1$ grade element of a multivector with grades $s - r + 1$ and $s + r - 1$, which has no terms for $r > 0$. Now we can expand the $A_s b$ multivector product, for
\begin{aligned}A_s \cdot \left( { b \wedge B_r } \right)={\left\langle{{ \left( { A_s \cdot b + A_s \wedge b} \right) B_r }}\right\rangle}_{{s - (r + 1)}}.\end{aligned} \hspace{\stretch{1}}(1.161)
The latter multivector (with the wedge product factor) above has grades $s + 1 - r$ and $s + 1 + r$, so this selection operator finds nothing. This leaves
\begin{aligned}A_s \cdot \left( { b \wedge B_r } \right)={\left\langle{{\left( { A_s \cdot b } \right) \cdot B_r+ \left( { A_s \cdot b } \right) \wedge B_r}}\right\rangle}_{{s - (r + 1)}}.\end{aligned} \hspace{\stretch{1}}(1.162)
The first dot products term has grade $s - 1 - r$ and is selected, whereas the wedge term has grade $s - 1 + r \ne s - r - 1$ (for $r > 0$). $\qquad\square$
### Lemma 2. Distribution of two bivectors
For vectors $\mathbf{a}$, $\mathbf{b}$, and bivector $B$, we have
\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b} } \right) \cdot B = \frac{1}{2} \left( {\mathbf{a} \cdot \left( { \mathbf{b} \cdot B } \right)-\mathbf{b} \cdot \left( { \mathbf{a} \cdot B } \right)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.163)
Proof follows by applying the scalar selection operator, expanding the wedge product within it, and eliminating any of the terms that cannot contribute grade zero values
\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b} } \right) \cdot B &= \left\langle{{\frac{1}{2} \Bigl( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \Bigr) B}}\right\rangle \\ &= \frac{1}{2}\left\langle{{\mathbf{a} \left( { \mathbf{b} \cdot B + \not{{ \mathbf{b} \wedge B }} } \right)-\mathbf{b} \left( { \mathbf{a} \cdot B + \not{{ \mathbf{a} \wedge B }} } \right)}}\right\rangle \\ &= \frac{1}{2}\left\langle{{\mathbf{a} \cdot \left( { \mathbf{b} \cdot B } \right)+\not{{\mathbf{a} \wedge \left( { \mathbf{b} \cdot B } \right)}}-\mathbf{b} \cdot \left( { \mathbf{a} \cdot B } \right)-\not{{\mathbf{b} \wedge \left( { \mathbf{a} \cdot B } \right)}}}}\right\rangle \\ &= \frac{1}{2}\Bigl( {\mathbf{a} \cdot \left( { \mathbf{b} \cdot B } \right)-\mathbf{b} \cdot \left( { \mathbf{a} \cdot B } \right)} \Bigr)\qquad\square\end{aligned} \hspace{\stretch{1}}(1.0.163)
### Lemma 3. Inner product of trivector with bivector
Given a bivector $B$, and trivector $\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}$ where $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are vectors, the inner product is
\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} } \right) \cdot B=\mathbf{a} \Bigl( { \left( { \mathbf{b} \wedge \mathbf{c} } \right) \cdot B } \Bigr)+\mathbf{b} \Bigl( { \left( { \mathbf{c} \wedge \mathbf{a} } \right) \cdot B } \Bigr)+\mathbf{c} \Bigl( { \left( { \mathbf{a} \wedge \mathbf{b} } \right) \cdot B } \Bigr).\end{aligned} \hspace{\stretch{1}}(1.165)
This is also problem 1.1(c) from Exercises 2.1 in [3], and submits to a dumb expansion in successive dot products with a final regrouping. With $B = \mathbf{u} \wedge \mathbf{v}$
\begin{aligned}\begin{aligned}\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right)\cdot B&={\left\langle{{\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \left( \mathbf{u} \wedge \mathbf{v} \right) }}\right\rangle}_{1} \\ &={\left\langle{{\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right)\left(\mathbf{u} \mathbf{v}- \mathbf{u} \cdot \mathbf{v}\right) }}\right\rangle}_{1} \\ &=\left(\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \cdot \mathbf{u} \right) \cdot \mathbf{v} \\ &=\left( \mathbf{a} \wedge \mathbf{b} \right) \cdot \mathbf{v} \left( \mathbf{c} \cdot \mathbf{u} \right)+\left( \mathbf{c} \wedge \mathbf{a} \right) \cdot \mathbf{v} \left( \mathbf{b} \cdot \mathbf{u} \right)+\left( \mathbf{b} \wedge \mathbf{c} \right) \cdot \mathbf{v} \left( \mathbf{a} \cdot \mathbf{u} \right) \\ &=\mathbf{a}\left( \mathbf{b} \cdot \mathbf{v} \right)\left( \mathbf{c} \cdot \mathbf{u} \right)-\mathbf{b}\left( \mathbf{a} \cdot \mathbf{v} \right)\left( \mathbf{c} \cdot \mathbf{u} \right) \\ &\quad +\mathbf{c}\left( \mathbf{a} \cdot \mathbf{v} \right)\left( \mathbf{b} \cdot \mathbf{u} \right)-\mathbf{a}\left( \mathbf{c} \cdot \mathbf{v} \right)\left( \mathbf{b} \cdot \mathbf{u} \right) \\ &\quad +\mathbf{b}\left( \mathbf{c} \cdot \mathbf{v} \right)\left( \mathbf{a} \cdot \mathbf{u} \right)-\mathbf{c}\left( \mathbf{b} \cdot \mathbf{v} \right)\left( \mathbf{a} \cdot \mathbf{u} \right) \\ &=\mathbf{a}\left( \left( \mathbf{b} \cdot \mathbf{v} \right) \left( \mathbf{c} \cdot \mathbf{u} \right) - \left( \mathbf{c} \cdot \mathbf{v} \right) \left( \mathbf{b} \cdot \mathbf{u} \right) \right)\\ &\quad +\mathbf{b}\left( \left( \mathbf{c} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right) - \left( \mathbf{a} \cdot \mathbf{v} \right) \left( \mathbf{c} \cdot \mathbf{u} \right) \right)\\ &\quad +\mathbf{c}\left( \left( \mathbf{a} \cdot \mathbf{v} \right) \left( \mathbf{b} \cdot \mathbf{u} \right) - \left( \mathbf{b} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right) \right) \\ &=\mathbf{a}\left( \mathbf{b} \wedge \mathbf{c} \right)\cdot\left( \mathbf{u} \wedge \mathbf{v} \right)\\ &\quad +\mathbf{b}\left( \mathbf{c} \wedge \mathbf{a} \right)\cdot\left( \mathbf{u} \wedge \mathbf{v} \right)\\ &\quad +\mathbf{c}\left( \mathbf{a} \wedge \mathbf{b} \right) \cdot\left( \mathbf{u} \wedge \mathbf{v} \right)\\ &=\mathbf{a}\left( \mathbf{b} \wedge \mathbf{c} \right)\cdot B+\mathbf{b}\left( \mathbf{c} \wedge \mathbf{a} \right) \cdot B+\mathbf{c}\left( \mathbf{a} \wedge \mathbf{b} \right)\cdot B. \qquad\square\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.166)
### Lemma 4. Distribution of two trivectors
Given a trivector $T$ and three vectors $\mathbf{a}, \mathbf{b}$, and $\mathbf{c}$, the entire inner product can be expanded in terms of any successive set inner products, subject to change of sign with interchange of any two adjacent vectors within the dot product sequence
\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} } \right) \cdot T &= \mathbf{a} \cdot \left( { \mathbf{b} \cdot \left( { \mathbf{c} \cdot T } \right) } \right) \\ &= -\mathbf{a} \cdot \left( { \mathbf{c} \cdot \left( { \mathbf{b} \cdot T } \right) } \right) \\ &= \mathbf{b} \cdot \left( { \mathbf{c} \cdot \left( { \mathbf{a} \cdot T } \right) } \right) \\ &= - \mathbf{b} \cdot \left( { \mathbf{a} \cdot \left( { \mathbf{c} \cdot T } \right) } \right) \\ &= \mathbf{c} \cdot \left( { \mathbf{a} \cdot \left( { \mathbf{b} \cdot T } \right) } \right) \\ &= - \mathbf{c} \cdot \left( { \mathbf{b} \cdot \left( { \mathbf{a} \cdot T } \right) } \right).\end{aligned} \hspace{\stretch{1}}(1.167)
To show this, we first expand within a scalar selection operator
\begin{aligned}\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} } \right) \cdot T&=\left\langle{{\left( { \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} } \right) T}}\right\rangle \\ &=\frac{1}{6}\left\langle{{ \mathbf{a} \mathbf{b} \mathbf{c} T- \mathbf{a} \mathbf{c} \mathbf{b} T+ \mathbf{b} \mathbf{c} \mathbf{a} T- \mathbf{b} \mathbf{a} \mathbf{b} T+ \mathbf{c} \mathbf{a} \mathbf{b} T- \mathbf{c} \mathbf{b} \mathbf{a} T}}\right\rangle \\ \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.168)
Now consider any single term from the scalar selection, such as the first. This can be reordered using the vector dot product identity
\begin{aligned}\left\langle{{ \mathbf{a} \mathbf{b} \mathbf{c} T}}\right\rangle=\left\langle{{ \mathbf{a} \left( { -\mathbf{c} \mathbf{b} + 2 \mathbf{b} \cdot \mathbf{c} } \right) T}}\right\rangle=-\left\langle{{ \mathbf{a} \mathbf{c} \mathbf{b} T}}\right\rangle+2 \mathbf{b} \cdot \mathbf{c} \not{{\left\langle{{ \mathbf{a} T}}\right\rangle}}.\end{aligned} \hspace{\stretch{1}}(1.0.168)
The vector-trivector product in the latter grade selection operation above contributes only bivector and quadvector terms, thus contributing nothing. This can be repeated, showing that
\begin{aligned} \left\langle{{ \mathbf{a} \mathbf{b} \mathbf{c} T }}\right\rangle &= - \left\langle{{ \mathbf{a} \mathbf{c} \mathbf{b} T }}\right\rangle \\ &= + \left\langle{{ \mathbf{b} \mathbf{c} \mathbf{a} T }}\right\rangle \\ &= - \left\langle{{ \mathbf{b} \mathbf{a} \mathbf{c} T }}\right\rangle \\ &= + \left\langle{{ \mathbf{c} \mathbf{a} \mathbf{b} T }}\right\rangle \\ &= - \left\langle{{ \mathbf{c} \mathbf{b} \mathbf{a} T }}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.0.168)
Substituting this back into eq. 1.0.168 proves lemma 4.
### Lemma 5. Permutation of two successive dot products with trivector
Given a trivector $T$ and two vectors $\mathbf{a}$ and $\mathbf{b}$, alternating the order of the dot products changes the sign
\begin{aligned}\mathbf{a} \cdot \left( { \mathbf{b} \cdot T } \right)=-\mathbf{b} \cdot \left( { \mathbf{a} \cdot T } \right).\end{aligned} \hspace{\stretch{1}}(1.171)
This and lemma 4 are clearly examples of a more general identity, but I’ll not try to prove that here. To show this one, we have
\begin{aligned}\mathbf{a} \cdot \left( { \mathbf{b} \cdot T } \right) &= {\left\langle{{ \mathbf{a} \left( { \mathbf{b} \cdot T } \right) }}\right\rangle}_{1} \\ &= \frac{1}{2}{\left\langle{{ \mathbf{a} \mathbf{b} T + \mathbf{a} T \mathbf{b} }}\right\rangle}_{1} \\ &= \frac{1}{2}{\left\langle{{ \left( { -\mathbf{b} \mathbf{a} + \not{{2 \mathbf{a} \cdot \mathbf{b}}}} \right) T + \left( { \mathbf{a} \cdot T} \right) \mathbf{b} + \not{{ \mathbf{a} \wedge T}} \mathbf{b} }}\right\rangle}_{1} \\ &= \frac{1}{2}\left( {-\mathbf{b} \cdot \left( { \mathbf{a} \cdot T } \right)+\left( { \mathbf{a} \cdot T } \right) \cdot \mathbf{b}} \right) \\ &= -\mathbf{b} \cdot \left( { \mathbf{a} \cdot T } \right). \qquad\square\end{aligned} \hspace{\stretch{1}}(1.172)
Cancellation of terms above was because they could not contribute to a grade one selection. We also employed the relation $\mathbf{x} \cdot B = - B \cdot \mathbf{x}$ for bivector $B$ and vector $\mathbf{x}$.
### Lemma 6. Duality in a plane
For a vector $\mathbf{a}$, and a plane containing $\mathbf{a}$ and $\mathbf{b}$, the dual $\mathbf{a}^{*}$ of this vector with respect to this plane is
\begin{aligned}\mathbf{a}^{*} = \frac{\mathbf{b} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right)}{\left( {\mathbf{a} \wedge \mathbf{b}} \right)^2},\end{aligned} \hspace{\stretch{1}}(1.173)
Satisfying
\begin{aligned}\mathbf{a}^{*} \cdot \mathbf{a} = 1,\end{aligned} \hspace{\stretch{1}}(1.174)
and
\begin{aligned}\mathbf{a}^{*} \cdot \mathbf{b} = 0.\end{aligned} \hspace{\stretch{1}}(1.175)
\begin{aligned}\mathbf{b} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right)=\left( { \mathbf{b} \cdot \mathbf{a} } \right) \mathbf{b}-\mathbf{b}^2 \mathbf{a}.\end{aligned} \hspace{\stretch{1}}(1.176)
Dotting with $\mathbf{a}$ we have
\begin{aligned}\mathbf{a} \cdot \left( { \mathbf{b} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right) } \right)=\mathbf{a} \cdot \left( {\left( { \mathbf{b} \cdot \mathbf{a} } \right) \mathbf{b}-\mathbf{b}^2 \mathbf{a}} \right)=\left( { \mathbf{b} \cdot \mathbf{a} } \right)^2 - \mathbf{b}^2 \mathbf{a}^2,\end{aligned} \hspace{\stretch{1}}(1.177)
but dotting with $\mathbf{b}$ yields zero
\begin{aligned}\mathbf{b} \cdot \left( { \mathbf{b} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right) } \right) &= \mathbf{b} \cdot \left( {\left( { \mathbf{b} \cdot \mathbf{a} } \right) \mathbf{b}-\mathbf{b}^2 \mathbf{a}} \right) \\ &= \left( { \mathbf{b} \cdot \mathbf{a} } \right) \mathbf{b}^2 - \mathbf{b}^2 \left( { \mathbf{a} \cdot \mathbf{b} } \right) \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.178)
To complete the proof, we note that the product in eq. 1.177 is just the wedge squared
\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b}} \right)^2 &= \left\langle{{\left( { \mathbf{a} \wedge \mathbf{b} } \right)^2}}\right\rangle \\ &= \left\langle{{\left( { \mathbf{a} \mathbf{b} - \mathbf{a} \cdot \mathbf{b} } \right)\left( { \mathbf{a} \mathbf{b} - \mathbf{a} \cdot \mathbf{b} } \right)}}\right\rangle \\ &= \left\langle{{\mathbf{a} \mathbf{b} \mathbf{a} \mathbf{b} - 2 \left( {\mathbf{a} \cdot \mathbf{b}} \right) \mathbf{a} \mathbf{b}}}\right\rangle+\left( { \mathbf{a} \cdot \mathbf{b} } \right)^2 \\ &= \left\langle{{\mathbf{a} \mathbf{b} \left( { -\mathbf{b} \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} } \right)}}\right\rangle-\left( { \mathbf{a} \cdot \mathbf{b} } \right)^2 \\ &= \left( { \mathbf{a} \cdot \mathbf{b} } \right)^2-\mathbf{a}^2 \mathbf{b}^2.\end{aligned} \hspace{\stretch{1}}(1.179)
This duality relation can be recast with a linear denominator
\begin{aligned}\mathbf{a}^{*} &= \frac{\mathbf{b} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right)}{\left( {\mathbf{a} \wedge \mathbf{b}} \right)^2} \\ &= \mathbf{b} \frac{\mathbf{a} \wedge \mathbf{b} }{\left( {\mathbf{a} \wedge \mathbf{b}} \right)^2} \\ &= \mathbf{b} \frac{\mathbf{a} \wedge \mathbf{b} }{\left\lvert {\mathbf{a} \wedge \mathbf{b} } \right\rvert} \frac{\left\lvert {\mathbf{a} \wedge \mathbf{b}} \right\rvert}{\mathbf{a} \wedge \mathbf{b} }\frac{1}{{\left( {\mathbf{a} \wedge \mathbf{b}} \right)}},\end{aligned} \hspace{\stretch{1}}(1.180)
or
\begin{aligned}\mathbf{a}^{*} = \mathbf{b} \frac{1}{{\left( {\mathbf{a} \wedge \mathbf{b}} \right)}}.\end{aligned} \hspace{\stretch{1}}(1.0.181)
We can use this form after scaling it appropriately to express duality in terms of the pseudoscalar.
### Lemma 7. Dual vector in a three vector subspace
In the subspace spanned by $\left\{ {\mathbf{a}, \mathbf{b}, \mathbf{c}} \right\}$, the dual of $\mathbf{a}$ is
\begin{aligned}\mathbf{a}^{*} = \mathbf{b} \wedge \mathbf{c} \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}},\end{aligned}
Consider the dot product of $\hat{\mathbf{a}}^{*}$ with $\mathbf{u} \in \left\{ {\mathbf{a}, \mathbf{b}, \mathbf{c}} \right\}$.
\begin{aligned}\mathbf{u} \cdot \mathbf{a}^{*} &= \left\langle{{ \mathbf{u} \mathbf{b} \wedge \mathbf{c} \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}} }}\right\rangle \\ &= \left\langle{{ \mathbf{u} \cdot \left( { \mathbf{b} \wedge \mathbf{c}} \right) \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}} }}\right\rangle+\left\langle{{ \mathbf{u} \wedge \mathbf{b} \wedge \mathbf{c} \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}} }}\right\rangle \\ &= \not{{\left\langle{{ \left( { \left( { \mathbf{u} \cdot \mathbf{b}} \right) \mathbf{c}-\left( {\mathbf{u} \cdot \mathbf{c}} \right) \mathbf{b}} \right)\frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}} }}\right\rangle}}+\left\langle{{ \mathbf{u} \wedge \mathbf{b} \wedge \mathbf{c} \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}} }}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.182)
The canceled term is eliminated since it is the product of a vector and trivector producing no scalar term. Substituting $\mathbf{a}, \mathbf{b}, \mathbf{c}$, and noting that $\mathbf{u} \wedge \mathbf{u} = 0$, we have
\begin{aligned}\begin{aligned}\mathbf{a} \cdot \mathbf{a}^{*} &= 1 \\ \mathbf{b} \cdot \mathbf{a}^{*} &= 0 \\ \mathbf{c} \cdot \mathbf{a}^{*} &= 0.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.183)
### Lemma 8. Pseudoscalar selection
For grade $k$ blade $K \in \bigwedge^k$ (i.e. a pseudoscalar), and vectors $\mathbf{a}, \mathbf{b}$, the grade $k$ selection of this blade sandwiched between the vectors is
\begin{aligned}{\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k} = (-1)^{k+1} {\left\langle{{K a b}}\right\rangle}_{k} = (-1)^{k+1} K \left( { \mathbf{a} \cdot \mathbf{b}} \right).\end{aligned}
To show this, we have to consider even and odd grades separately. First for even $k$ we have
\begin{aligned}{\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k} &= {\left\langle{{ \left( { \mathbf{a} \cdot K + \not{{\mathbf{a} \wedge K}}} \right) \mathbf{b} }}\right\rangle}_{k} \\ &= \frac{1}{2} {\left\langle{{ \left( { \mathbf{a} K - K \mathbf{a} } \right) \mathbf{b} }}\right\rangle}_{k} \\ &= \frac{1}{2} {\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k}-\frac{1}{2} {\left\langle{{ K \mathbf{a} \mathbf{b} }}\right\rangle}_{k},\end{aligned} \hspace{\stretch{1}}(1.184)
or
\begin{aligned}{\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k} = -{\left\langle{{ K \mathbf{a} \mathbf{b} }}\right\rangle}_{k} = -K \left( { \mathbf{a} \cdot \mathbf{b}} \right).\end{aligned} \hspace{\stretch{1}}(1.185)
Similarly for odd $k$, we have
\begin{aligned}{\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k} &= {\left\langle{{ \left( { \mathbf{a} \cdot K + \not{{\mathbf{a} \wedge K}}} \right) \mathbf{b} }}\right\rangle}_{k} \\ &= \frac{1}{2} {\left\langle{{ \left( { \mathbf{a} K + K \mathbf{a} } \right) \mathbf{b} }}\right\rangle}_{k} \\ &= \frac{1}{2} {\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k}+\frac{1}{2} {\left\langle{{ K \mathbf{a} \mathbf{b} }}\right\rangle}_{k},\end{aligned} \hspace{\stretch{1}}(1.186)
or
\begin{aligned}{\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k} = {\left\langle{{ K \mathbf{a} \mathbf{b} }}\right\rangle}_{k} = K \left( { \mathbf{a} \cdot \mathbf{b}} \right).\end{aligned} \hspace{\stretch{1}}(1.187)
Adjusting for the signs completes the proof.
# References
[1] John Denker. Magnetic field for a straight wire., 2014. URL http://www.av8n.com/physics/straight-wire.pdf. [Online; accessed 11-May-2014].
[2] H. Flanders. Differential Forms With Applications to the Physical Sciences. Courier Dover Publications, 1989.
[3] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.
[4] Peeter Joot. Collection of old notes on Stokes theorem in Geometric algebra, 2014. URL https://sites.google.com/site/peeterjoot3/math2014/bigCollectionOfPartiallyIncorrectStokesTheoremMusings.pdf.
[5] Peeter Joot. Synposis of old notes on Stokes theorem in Geometric algebra, 2014. URL https://sites.google.com/site/peeterjoot3/math2014/synopsisOfBigCollectionOfPartiallyIncorrectStokesTheoremMusings.pdf.
[6] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.
[7] M. Schwartz. Principles of Electrodynamics. Dover Publications, 1987.
[8] Michael Spivak. Calculus on manifolds, volume 1. Benjamin New York, 1965.
## Geometric Algebra. The very quickest introduction.
Posted by peeterjoot on March 17, 2012
# Motivation.
An attempt to make a relatively concise introduction to Geometric (or Clifford) Algebra. Much more complete introductions to the subject can be found in [1], [2], and [3].
# Axioms
We have a couple basic principles upon which the algebra is based
1. Vectors can be multiplied.
2. The square of a vector is the (squared) length of that vector (with appropriate generalizations for non-Euclidean metrics).
3. Vector products are associative (but not necessarily commutative).
That’s really all there is to it, and the rest, paraphrasing Feynman, can be figured out by anybody sufficiently clever.
# By example. The 2D case.
Consider a 2D Euclidean space, and the product of two vectors $\mathbf{a}$ and $\mathbf{b}$ in that space. Utilizing a standard orthonormal basis $\{\mathbf{e}_1, \mathbf{e}_2\}$ we can write
\begin{aligned}\mathbf{a} &= \mathbf{e}_1 x_1 + \mathbf{e}_2 x_2 \\ \mathbf{b} &= \mathbf{e}_1 y_1 + \mathbf{e}_2 y_2,\end{aligned} \hspace{\stretch{1}}(3.1)
and let’s write out the product of these two vectors $\mathbf{a} \mathbf{b}$, not yet knowing what we will end up with. That is
\begin{aligned}\mathbf{a} \mathbf{b} &= (\mathbf{e}_1 x_1 + \mathbf{e}_2 x_2 )( \mathbf{e}_1 y_1 + \mathbf{e}_2 y_2 ) \\ &= \mathbf{e}_1^2 x_1 y_1 + \mathbf{e}_2^2 x_2 y_2+ \mathbf{e}_1 \mathbf{e}_2 x_1 y_2 + \mathbf{e}_2 \mathbf{e}_1 x_2 y_1\end{aligned}
From axiom 2 we have $\mathbf{e}_1^2 = \mathbf{e}_2^2 = 1$, so we have
\begin{aligned}\mathbf{a} \mathbf{b} = x_1 y_1 + x_2 y_2 + \mathbf{e}_1 \mathbf{e}_2 x_1 y_2 + \mathbf{e}_2 \mathbf{e}_1 x_2 y_1.\end{aligned} \hspace{\stretch{1}}(3.3)
We’ve multiplied two vectors and ended up with a scalar component (and recognize that this part of the vector product is the dot product), and a component that is a “something else”. We’ll call this something else a bivector, and see that it is characterized by a product of non-colinear vectors. These products $\mathbf{e}_1 \mathbf{e}_2$ and $\mathbf{e}_2 \mathbf{e}_1$ are in fact related, and we can see that by looking at the case of $\mathbf{b} = \mathbf{a}$. For that we have
\begin{aligned}\mathbf{a}^2 &=x_1 x_1 + x_2 x_2 + \mathbf{e}_1 \mathbf{e}_2 x_1 x_2 + \mathbf{e}_2 \mathbf{e}_1 x_2 x_1 \\ &={\left\lvert{\mathbf{a}}\right\rvert}^2 +x_1 x_2 ( \mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_2 \mathbf{e}_1 )\end{aligned}
Since axiom (2) requires our vectors square to equal its (squared) length, we must then have
\begin{aligned}\mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_2 \mathbf{e}_1 = 0,\end{aligned} \hspace{\stretch{1}}(3.4)
or
\begin{aligned}\mathbf{e}_2 \mathbf{e}_1 = -\mathbf{e}_1 \mathbf{e}_2.\end{aligned} \hspace{\stretch{1}}(3.5)
We see that Euclidean orthonormal vectors anticommute. What we can see with some additional study is that any colinear vectors commute, and in Euclidean spaces (of any dimension) vectors that are normal to each other anticommute (this can also be taken as a definition of normal).
We can now return to our product of two vectors 3.3 and simplify it slightly
\begin{aligned}\mathbf{a} \mathbf{b} = x_1 y_1 + x_2 y_2 + \mathbf{e}_1 \mathbf{e}_2 (x_1 y_2 - x_2 y_1).\end{aligned} \hspace{\stretch{1}}(3.6)
The product of two vectors in 2D is seen here to have one scalar component, and one bivector component (an irreducible product of two normal vectors). Observe the symmetric and antisymmetric split of the scalar and bivector components above. This symmetry and antisymmetry can be made explicit, introducing dot and wedge product notation respectively
\begin{aligned}\mathbf{a} \cdot \mathbf{b} &= \frac{1}{{2}}( \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a}) = x_1 y_1 + x_2 y_2 \\ \mathbf{a} \wedge \mathbf{b} &= \frac{1}{{2}}( \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a}) = \mathbf{e}_1 \mathbf{e}_2 (x_1 y_y - x_2 y_1).\end{aligned} \hspace{\stretch{1}}(3.7)
so that the vector product can be written as
\begin{aligned}\mathbf{a} \mathbf{b} = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \wedge \mathbf{b}.\end{aligned} \hspace{\stretch{1}}(3.9)
# Pseudoscalar
In many contexts it is useful to introduce an ordered product of all the unit vectors for the space is called the pseudoscalar. In our 2D case this is
\begin{aligned}i = \mathbf{e}_1 \mathbf{e}_2,\end{aligned} \hspace{\stretch{1}}(4.10)
a quantity that we find behaves like the complex imaginary. That can be shown by considering its square
\begin{aligned}(\mathbf{e}_1 \mathbf{e}_2)^2&=(\mathbf{e}_1 \mathbf{e}_2)(\mathbf{e}_1 \mathbf{e}_2) \\ &=\mathbf{e}_1 (\mathbf{e}_2 \mathbf{e}_1) \mathbf{e}_2 \\ &=-\mathbf{e}_1 (\mathbf{e}_1 \mathbf{e}_2) \mathbf{e}_2 \\ &=-(\mathbf{e}_1 \mathbf{e}_1) (\mathbf{e}_2 \mathbf{e}_2) \\ &=-1^2 \\ &= -1\end{aligned}
Here the anticommutation of normal vectors property has been used, as well as (for the first time) the associative multiplication axiom.
In a 3D context, you’ll see the pseudoscalar in many places (expressing the normals to planes for example). It also shows up in a number of fundamental relationships. For example, if one writes
\begin{aligned}I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(4.11)
for the 3D pseudoscalar, then it’s also possible to show
\begin{aligned}\mathbf{a} \mathbf{b} = \mathbf{a} \cdot \mathbf{b} + I (\mathbf{a} \times \mathbf{b})\end{aligned} \hspace{\stretch{1}}(4.12)
something that will be familiar to the student of QM, where we see this in the context of Pauli matrices. The Pauli matrices also encode a Clifford algebraic structure, but we do not need an explicit matrix representation to do so.
# Rotations
Very much like complex numbers we can utilize exponentials to perform rotations. Rotating in a sense from $\mathbf{e}_1$ to $\mathbf{e}_2$, can be expressed as
\begin{aligned}\mathbf{a} e^{i \theta}&=(\mathbf{e}_1 x_1 + \mathbf{e}_2 x_2) (\cos\theta + \mathbf{e}_1 \mathbf{e}_2 \sin\theta) \\ &=\mathbf{e}_1 (x_1 \cos\theta - x_2 \sin\theta)+\mathbf{e}_2 (x_2 \cos\theta + x_1 \sin\theta)\end{aligned}
More generally, even in N dimensional Euclidean spaces, if $\mathbf{a}$ is a vector in a plane, and $\hat{\mathbf{u}}$ and $\hat{\mathbf{v}}$ are perpendicular unit vectors in that plane, then the rotation through angle $\theta$ is given by
\begin{aligned}\mathbf{a} \rightarrow \mathbf{a} e^{\hat{\mathbf{u}} \hat{\mathbf{v}} \theta}.\end{aligned} \hspace{\stretch{1}}(5.13)
This is illustrated in figure (1).
Plane rotation.
Notice that we have expressed the rotation here without utilizing a normal direction for the plane. The sense of the rotation is encoded by the bivector $\hat{\mathbf{u}} \hat{\mathbf{v}}$ that describes the plane and the orientation of the rotation (or by duality the direction of the normal in a 3D space). By avoiding a requirement to encode the rotation using a normal to the plane we have an method of expressing the rotation that works not only in 3D spaces, but also in 2D and greater than 3D spaces, something that isn’t possible when we restrict ourselves to traditional vector algebra (where quantities like the cross product can’t be defined in a 2D or 4D space, despite the fact that things they may represent, like torque are planar phenomena that do not have any intrinsic requirement for a normal that falls out of the plane.).
When $\mathbf{a}$ does not lie in the plane spanned by the vectors $\hat{\mathbf{u}}$ and $\hat{\mathbf{v}}$ , as in figure (2), we must express the rotations differently. A rotation then takes the form
\begin{aligned}\mathbf{a} \rightarrow e^{-\hat{\mathbf{u}} \hat{\mathbf{v}} \theta/2} \mathbf{a} e^{\hat{\mathbf{u}} \hat{\mathbf{v}} \theta/2}.\end{aligned} \hspace{\stretch{1}}(5.14)
3D rotation.
In the 2D case, and when the vector lies in the plane this reduces to the one sided complex exponential operator used above. We see these types of paired half angle rotations in QM, and they are also used extensively in computer graphics under the guise of quaternions.
# References
[1] L. Dorst, D. Fontijne, and S. Mann. Geometric Algebra for Computer Science. Morgan Kaufmann, San Francisco, 2007.
[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.
[3] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.
## PHY456H1F: Quantum Mechanics II. Lecture 15 (Taught by Prof J.E. Sipe). Rotation operator in spin space
Posted by peeterjoot on October 31, 2011
[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]
# Disclaimer.
Peeter’s lecture notes from class. May not be entirely coherent.
# Rotation operator in spin space.
We can formally expand our rotation operator in Taylor series
\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar}= I +\left(-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar\right)+\frac{1}{{2!}}\left(-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar\right)^2+\frac{1}{{3!}}\left(-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar\right)^3+ \cdots\end{aligned} \hspace{\stretch{1}}(2.1)
or
\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2}&= I +\left(-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2\right)+\frac{1}{{2!}}\left(-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2\right)^2+\frac{1}{{3!}}\left(-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2\right)^3+ \cdots \\ &=\sigma_0 +\left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})+\frac{1}{{2!}} \left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^2+\frac{1}{{3!}} \left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^3+ \cdots \\ &=\sigma_0 +\left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})+\frac{1}{{2!}} \left(\frac{-i \theta}{2}\right) \sigma_0+\frac{1}{{3!}} \left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma}) + \cdots \\ &=\sigma_0 \left( 1 - \frac{1}{{2!}}\left(\frac{\theta}{2}\right)^2 + \cdots \right) +(\hat{\mathbf{n}} \cdot \boldsymbol{\sigma}) \left( \frac{\theta}{2} - \frac{1}{{3!}}\left(\frac{\theta}{2}\right)^3 + \cdots \right) \\ &=\cos(\theta/2) \sigma_0 + \sin(\theta/2) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})\end{aligned}
where we’ve used the fact that $(\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^2 = \sigma_0$.
So our representation of the spin operator is
\begin{aligned}\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} &\rightarrow \cos(\theta/2) \sigma_0 + \sin(\theta/2) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma}) \\ &=\cos(\theta/2) \sigma_0 + \sin(\theta/2) \left(n_x \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} + n_y \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} + n_z \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \right) \\ &=\begin{bmatrix}\cos(\theta/2) -i n_z \sin(\theta/2) & -i (n_x -i n_y) \sin(\theta/2) \\ -i (n_x + i n_y) \sin(\theta/2) & \cos(\theta/2) +i n_z \sin(\theta/2) \end{bmatrix}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.2)
Note that, in particular,
\begin{aligned}e^{-2 \pi i \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} \rightarrow \cos\pi \sigma_0 = -\sigma_0\end{aligned} \hspace{\stretch{1}}(2.3)
This “rotates” the ket, but introduces a phase factor.
Can do this in general for other degrees of spin, for $s = 1/2, 3/2, 5/2, \cdots$.
## Unfortunate interjection by me
I mentioned the half angle rotation operator that requires a half angle operator sandwich. Prof. Sipe thought I might be talking about a Heisenberg picture representation, where we have something like this in expectation values
\begin{aligned}{\left\lvert {\psi'} \right\rangle} = e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.4)
so that
\begin{aligned}{\left\langle {\psi'} \right\rvert}\mathcal{O}{\left\lvert {\psi'} \right\rangle} = {\left\langle {\psi} \right\rvert} e^{i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} \mathcal{O}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.5)
However, what I was referring to, was that a general rotation of a vector in a Pauli matrix basis
\begin{aligned}R(\sum a_k \sigma_k) = R( \mathbf{a} \cdot \boldsymbol{\sigma})\end{aligned} \hspace{\stretch{1}}(2.6)
can be expressed by sandwiching the Pauli vector representation by two half angle rotation operators like our spin 1/2 operators from class today
\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = e^{-\theta \hat{\mathbf{u}} \cdot \boldsymbol{\sigma} \hat{\mathbf{v}} \cdot \boldsymbol{\sigma}/2} \mathbf{a} \cdot \boldsymbol{\sigma} e^{\theta \hat{\mathbf{u}} \cdot \boldsymbol{\sigma} \hat{\mathbf{v}} \cdot \boldsymbol{\sigma}/2}\end{aligned} \hspace{\stretch{1}}(2.7)
where $\hat{\mathbf{u}}$ and $\hat{\mathbf{v}}$ are two non-colinear orthogonal unit vectors that define the oriented plane that we are rotating in.
For example, rotating in the $x-y$ plane, with $\hat{\mathbf{u}} = \hat{\mathbf{x}}$ and $\hat{\mathbf{v}} = \hat{\mathbf{y}}$, we have
\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = e^{-\theta \sigma_1 \sigma_2/2} (a_1 \sigma_1 + a_2 \sigma_2 + a_3 \sigma_3) e^{\theta \sigma_1 \sigma_2/2} \end{aligned} \hspace{\stretch{1}}(2.8)
Observe that these exponentials commute with $\sigma_3$, leaving
\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) &= (a_1 \sigma_1 + a_2 \sigma_2) e^{\theta \sigma_1 \sigma_2} + a_3 \sigma_3 \\ &= (a_1 \sigma_1 + a_2 \sigma_2) (\cos\theta + \sigma_1 \sigma_2 \sin\theta)+a_3 \sigma_3 \\ &= \sigma_1 (a_1 \cos\theta - a_2 \sin\theta)+ \sigma_2 (a_2 \cos\theta + a_1 \sin\theta)+ \sigma_3 (a_3)\end{aligned}
yielding our usual coordinate rotation matrix. Expressed in terms of a unit normal to that plane, we form the normal by multiplication with the unit spatial volume element $I = \sigma_1 \sigma_2 \sigma_3$. For example:
\begin{aligned}\sigma_1 \sigma_2 \sigma_3( \sigma_3 )=\sigma_1 \sigma_2 \end{aligned} \hspace{\stretch{1}}(2.9)
and can in general write a spatial rotation in a Pauli basis representation as a sandwich of half angle rotation matrix exponentials
\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = e^{-I \theta (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})/2} (\mathbf{a} \cdot \boldsymbol{\sigma})e^{I \theta (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})/2} \end{aligned} \hspace{\stretch{1}}(2.10)
when $\hat{\mathbf{n}} \cdot \mathbf{a} = 0$ we get the complex-number like single sided exponential rotation exponentials (since $\mathbf{a} \cdot \boldsymbol{\sigma}$ commutes with $\mathbf{n} \cdot \boldsymbol{\sigma}$ in that case)
\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = (\mathbf{a} \cdot \boldsymbol{\sigma} )e^{I \theta (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})} \end{aligned} \hspace{\stretch{1}}(2.11)
I believe it was pointed out in one of [1] or [2] that rotations expressed in terms of half angle Pauli matrices has caused some confusion to students of quantum mechanics, because this $2 \pi$ “rotation” only generates half of the full spatial rotation. It was argued that this sort of confusion can be avoided if one observes that these half angle rotations exponentials are exactly what we require for general spatial rotations, and that a pair of half angle operators are required to produce a full spatial rotation.
The book [1] takes this a lot further, and produces a formulation of spin operators that is devoid of the normal scalar imaginary $i$ (using the Clifford algebra spatial unit volume element instead), and also does not assume a specific matrix representation of the spin operators. They argue that this leads to some subtleties associated with interpretation, but at the time I was attempting to read that text I did know enough QM to appreciate what they were doing, and haven’t had time to attempt a new study of that content.
# Spin dynamics
At least classically, the angular momentum of charged objects is associated with a magnetic moment as illustrated in figure (\ref{fig:qmTwoL15:qmTwoL15fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL15fig1}
\caption{Magnetic moment due to steady state current}
\end{figure}
\begin{aligned}\boldsymbol{\mu} = I A \mathbf{e}_\perp\end{aligned} \hspace{\stretch{1}}(3.12)
In our scheme, following the (cgs?) text conventions of [3], where the $\mathbf{E}$ and $\mathbf{B}$ have the same units, we write
\begin{aligned}\boldsymbol{\mu} = \frac{I A}{c} \mathbf{e}_\perp\end{aligned} \hspace{\stretch{1}}(3.13)
For a charge moving in a circle as in figure (\ref{fig:qmTwoL15:qmTwoL15fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL15fig2}
\caption{Charge moving in circle.}
\end{figure}
\begin{aligned}\begin{aligned}I &= \frac{\text{charge}}{\text{time}} \\ &= \frac{\text{distance}}{\text{time}} \frac{\text{charge}}{\text{distance}} \\ &= \frac{q v}{ 2 \pi r}\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.14)
so the magnetic moment is
\begin{aligned}\begin{aligned}\mu &= \frac{q v}{ 2 \pi r} \frac{\pi r^2}{c} \\ &= \frac{q }{ 2 m c } (m v r) \\ &= \gamma L\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.15)
Here $\gamma$ is the gyromagnetic ratio
Recall that we have a torque, as shown in figure (\ref{fig:qmTwoL15:qmTwoL15fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL15fig3}
\caption{Induced torque in the presence of a magnetic field.}
\end{figure}
\begin{aligned}\mathbf{T} = \boldsymbol{\mu} \times \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.16)
tending to line up $\boldsymbol{\mu}$ with $\mathbf{B}$. The energy is then
\begin{aligned}-\boldsymbol{\mu} \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.17)
Also recall that this torque leads to precession as shown in figure (\ref{fig:qmTwoL15:qmTwoL15fig4})
\begin{aligned}\frac{d{\mathbf{L}}}{dt} = \mathbf{T} = \gamma \mathbf{L} \times \mathbf{B},\end{aligned} \hspace{\stretch{1}}(3.18)
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL15fig4}
\caption{Precession due to torque.}
\end{figure}
with precession frequency
\begin{aligned}\boldsymbol{\omega} = - \gamma \mathbf{B}.\end{aligned} \hspace{\stretch{1}}(3.19)
For a current due to a moving electron
\begin{aligned}\gamma = -\frac{e}{2 m c} < 0\end{aligned} \hspace{\stretch{1}}(3.20)
where we are, here, writing for charge on the electron $-e$.
Question: steady state currents only?. Yes, this is only true for steady state currents.
For the translational motion of an electron, even if it is not moving in a steady way, regardless of it’s dynamics
\begin{aligned}\boldsymbol{\mu}_0 = - \frac{e}{2 m c} \mathbf{L}\end{aligned} \hspace{\stretch{1}}(3.21)
Now, back to quantum mechanics, we turn $\boldsymbol{\mu}_0$ into a dipole moment operator and $\mathbf{L}$ is “promoted” to an angular momentum operator.
\begin{aligned}H_{\text{int}} = - \boldsymbol{\mu}_0 \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.22)
Perhaps
\begin{aligned}\boldsymbol{\mu}_s = \gamma_s \mathbf{S}\end{aligned} \hspace{\stretch{1}}(3.23)
we write this as
\begin{aligned}\boldsymbol{\mu}_s = g \left( -\frac{e}{ 2 m c} \right)\mathbf{S}\end{aligned} \hspace{\stretch{1}}(3.24)
so that
\begin{aligned}\gamma_s = - \frac{g e}{ 2 m c} \end{aligned} \hspace{\stretch{1}}(3.25)
Experimentally, one finds to very good approximation
\begin{aligned}g = 2\end{aligned} \hspace{\stretch{1}}(3.26)
There was a lot of trouble with this in early quantum mechanics where people got things wrong, and canceled the wrong factors of $2$.
In fact, Dirac’s relativistic theory for the electron predicts $g=2$.
When this is measured experimentally, one does not get exactly $g=2$, and a theory that also incorporates photon creation and destruction and the interaction with the electron with such (virtual) photons. We get
\begin{aligned}\begin{aligned}g_{\text{theory}} &= 2 \left(1.001159652140 (\pm 28)\right) \\ g_{\text{experimental}} &= 2 \left(1.0011596521884 (\pm 43)\right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.27)
Richard Feynman compared the precision of quantum mechanics, referring to this measurement, “to predicting a distance as great as the width of North America to an accuracy of one human hair’s breadth”.
# References
[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.
[2] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.
[3] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.
## new version of geometric algebra notes compilation
Posted by peeterjoot on October 25, 2011
Dec 27, 2010 Vector form of Julia fractal.
Some typos are probably fixed too.
Changelog for version 9 was here.
Posted in geometric algebra | Tagged: , | 1 Comment »
## PHY456H1F: Quantum Mechanics II. Lecture 6 (Taught by Prof J.E. Sipe). Interaction picture.
Posted by peeterjoot on September 27, 2011
[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]
# Disclaimer.
Peeter’s lecture notes from class. May not be entirely coherent.
# Interaction picture.
## Recap.
Recall our table comparing our two interaction pictures
\begin{aligned}\text{Schr\"{o}dinger picture} &\qquad \text{Heisenberg picture} \\ i \hbar \frac{d}{dt} {\lvert {\psi_s(t)} \rangle} = H {\lvert {\psi_s(t)} \rangle} &\qquad i \hbar \frac{d}{dt} O_H(t) = \left[{O_H},{H}\right] \\ {\langle {\psi_s(t)} \rvert} O_S {\lvert {\psi_s(t)} \rangle} &= {\langle {\psi_H} \rvert} O_H {\lvert {\psi_H} \rangle} \\ {\lvert {\psi_s(0)} \rangle} &= {\lvert {\psi_H} \rangle} \\ O_S &= O_H(0)\end{aligned}
## A motivating example.
While fundamental Hamiltonians are independent of time, in a number of common cases, we can form approximate Hamiltonians that are time dependent. One such example is that of Coulomb excitations of an atom, as covered in section 18.3 of the text [1], and shown in figure (\ref{fig:qmTwoL6fig1}).
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL6fig1}
\caption{Coulomb interaction of a nucleus and heavy atom.}
\end{figure}
We consider the interaction of a nucleus with a neutral atom, heavy enough that it can be considered classically. From the atoms point of view, the effects of the heavy nucleus barreling by can be described using a time dependent Hamiltonian. For the atom, that interaction Hamiltonian is
\begin{aligned}H' = \sum_i \frac{ Z e q_i }{{\left\lvert{\mathbf{r}_N(t) - \mathbf{R}_i}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(2.1)
Here and $\mathbf{r}_N$ is the position vector for the heavy nucleus, and $\mathbf{R}_i$ is the position to each charge within the atom, where $i$ ranges over all the internal charges, positive and negative, within the atom.
Placing the origin close to the atom, we can write this interaction Hamiltonian as
\begin{aligned}H'(t) = \not{{\sum_i \frac{Z e q_i}{{\left\lvert{\mathbf{r}_N(t)}\right\rvert}}}}+ \sum_i Z e q_i \mathbf{R}_i \cdot {\left.{{\left(\frac{\partial {}}{\partial {\mathbf{r}}} \frac{1}{{{\left\lvert{ \mathbf{r}_N(t) - \mathbf{r}}\right\rvert}}}\right)}}\right\vert}_{{\mathbf{r} = 0}}\end{aligned} \hspace{\stretch{1}}(2.2)
The first term vanishes because the total charge in our neutral atom is zero. This leaves us with
\begin{aligned}\begin{aligned}H'(t) &= -\sum_i q_i \mathbf{R}_i \cdot {\left.{{\left(-\frac{\partial {}}{\partial {\mathbf{r}}} \frac{ Z e}{{\left\lvert{ \mathbf{r}_N(t) - \mathbf{r}}\right\rvert}}\right)}}\right\vert}_{{\mathbf{r} = 0}} \\ &= - \sum_i q_i \mathbf{R}_i \cdot \mathbf{E}(t),\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.3)
where $\mathbf{E}(t)$ is the electric field at the origin due to the nucleus.
Introducing a dipole moment operator for the atom
\begin{aligned}\boldsymbol{\mu} = \sum_i q_i \mathbf{R}_i,\end{aligned} \hspace{\stretch{1}}(2.4)
the interaction takes the form
\begin{aligned}H'(t) = -\boldsymbol{\mu} \cdot \mathbf{E}(t).\end{aligned} \hspace{\stretch{1}}(2.5)
Here we have a quantum mechanical operator, and a classical field taken together. This sort of dipole interaction also occurs when we treat a atom placed into an electromagnetic field, treated classically as depicted in figure (\ref{fig:qmTwoL6fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL6fig2}
\caption{atom in a field}
\end{figure}
In the figure, we can use the dipole interaction, provided $\lambda \gg a$, where $a$ is the “width” of the atom.
Because it is great for examples, we will see this dipole interaction a lot.
## The interaction picture.
Having talked about both the Schr\”{o}dinger and Heisenberg pictures, we can now move on to describe a hybrid, one where our Hamiltonian has been split into static and time dependent parts
\begin{aligned}H(t) = H_0 + H'(t)\end{aligned} \hspace{\stretch{1}}(2.6)
We will formulate an approach for dealing with problems of this sort called the interaction picture.
This is also covered in section 3.3 of the text, albeit in a much harder to understand fashion (the text appears to try to not pull the result from a magic hat, but the steps to get to the end result are messy). It would probably have been nicer to see it this way instead.
In the Schr\”{o}dinger picture our dynamics have the form
\begin{aligned}i \hbar \frac{d}{dt} {\lvert {\psi_s(t)} \rangle} = H {\lvert {\psi_s(t)} \rangle}\end{aligned} \hspace{\stretch{1}}(2.7)
How about the Heisenberg picture? We look for a solution
\begin{aligned}{\lvert {\psi_s(t)} \rangle} = U(t, t_0) {\lvert {\psi_s(t_0)} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.8)
We want to find this operator that evolves the state from the state as some initial time $t_0$, to the arbitrary later state found at time $t$. Plugging in we have
\begin{aligned}i \hbar \frac{d{{}}}{dt} U(t, t_0) {\lvert {\psi_s(t_0)} \rangle}=H(t) U(t, t_0) {\lvert {\psi_s(t_0)} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.9)
This has to hold for all ${\lvert {\psi_s(t_0)} \rangle}$, and we can equivalently seek a solution of the operator equation
\begin{aligned}i \hbar \frac{d{{}}}{dt} U(t, t_0) = H(t) U(t, t_0),\end{aligned} \hspace{\stretch{1}}(2.10)
where
\begin{aligned}U(t_0, t_0) = I,\end{aligned} \hspace{\stretch{1}}(2.11)
the identity for the Hilbert space.
Suppose that $H(t)$ was independent of time. We could find that
\begin{aligned}U(t, t_0) = e^{-i H(t - t_0)/\hbar}.\end{aligned} \hspace{\stretch{1}}(2.12)
If $H(t)$ depends on time could you guess that
\begin{aligned}U(t, t_0) = e^{-\frac{i}{\hbar} \int_{t_0}^t H(\tau) d\tau}\end{aligned} \hspace{\stretch{1}}(2.13)
holds? No. This may be true when $H(t)$ is a number, but when it is an operator, the Hamiltonian does not necessarily commute with itself at different times
\begin{aligned}\left[{H(t')},{H(t'')}\right] \ne 0.\end{aligned} \hspace{\stretch{1}}(2.14)
So this is wrong in general. As an aside, for numbers, 2.13 can be verified easily. We have
\begin{aligned}i \hbar \left( e^{-\frac{i}{\hbar} \int_{t_0}^t H(\tau) d\tau} \right)'&=i \hbar \left( -\frac{i}{\hbar} \right) \left( \int_{t_0}^t H(\tau) d\tau \right)'e^{-\frac{i}{\hbar} \int_{t_0}^t H(\tau) d\tau } \\ &=\left( H(t) \frac{dt}{dt} - H(t_0) \frac{dt_0}{dt} \right)e^{-\frac{i}{\hbar} \int_{t_0}^t H(\tau) d\tau} \\ &= H(t) U(t, t_0)\end{aligned}
## Expectations
Suppose that we do find $U(t, t_0)$. Then our expectation takes the form
\begin{aligned}{\langle {\psi_s(t)} \rvert} O_s {\lvert {\psi_s(t)} \rangle} = {\langle {\psi_s(t_0)} \rvert} U^\dagger(t, t_0) O_s U(t, t_0) {\lvert {\psi_s(t_0)} \rangle} \end{aligned} \hspace{\stretch{1}}(2.15)
Put
\begin{aligned}{\lvert {\psi_H} \rangle} = {\lvert {\psi_s(t_0)} \rangle},\end{aligned} \hspace{\stretch{1}}(2.16)
and form
\begin{aligned}O_H = U^\dagger(t, t_0) O_s U(t, t_0) \end{aligned} \hspace{\stretch{1}}(2.17)
so that our expectation has the familiar representations
\begin{aligned}{\langle {\psi_s(t)} \rvert} O_s {\lvert {\psi_s(t)} \rangle} ={\langle {\psi_H} \rvert} O_H {\lvert {\psi_H} \rangle} \end{aligned} \hspace{\stretch{1}}(2.18)
## New strategy. Interaction picture.
Let’s define
\begin{aligned}U_I(t, t_0) = e^{\frac{i}{\hbar} H_0(t - t_0)} U(t, t_0)\end{aligned} \hspace{\stretch{1}}(2.19)
or
\begin{aligned}U(t, t_0) = e^{-\frac{i}{\hbar} H_0(t - t_0)} U_I(t, t_0).\end{aligned} \hspace{\stretch{1}}(2.20)
Let’s see how this works. We have
\begin{aligned}i \hbar \frac{d{{U_I}}}{dt} &= i \hbar \frac{d{{}}}{dt} \left(e^{\frac{i}{\hbar} H_0(t - t_0)} U(t, t_0)\right) \\ &=-H_0 U(t, t_0)+e^{\frac{i}{\hbar} H_0(t - t_0)} \left( i \hbar \frac{d{{}}}{dt} U(t, t_0) \right) \\ &=-H_0 U(t, t_0)+e^{\frac{i}{\hbar} H_0(t - t_0)} \left( (H + H'(t)) U(t, t_0) \right) \\ &=e^{\frac{i}{\hbar} H_0(t - t_0)} H'(t)) U(t, t_0) \\ &=e^{\frac{i}{\hbar} H_0(t - t_0)} H'(t)) e^{-\frac{i}{\hbar} H_0(t - t_0)} U_I(t, t_0).\end{aligned}
Define
\begin{aligned}\bar{H}'(t) =e^{\frac{i}{\hbar} H_0(t - t_0)} H'(t)) e^{-\frac{i}{\hbar} H_0(t - t_0)},\end{aligned} \hspace{\stretch{1}}(2.21)
so that our operator equation takes the form
\begin{aligned}i \hbar \frac{d{{}}}{dt} U_I(t, t_0) = \bar{H}'(t) U_I(t, t_0).\end{aligned} \hspace{\stretch{1}}(2.22)
Note that we also have the required identity at the initial time
\begin{aligned}U_I(t_0, t_0) = I.\end{aligned} \hspace{\stretch{1}}(2.23)
Without requiring us to actually find $U(t, t_0)$ all of the dynamics of the time dependent interaction are now embedded in our operator equation for $\bar{H}'(t)$, with all of the simple interaction related to the non time dependent portions of the Hamiltonian left separate.
## Connection with the Schr\”{o}dinger picture.
In the Schr\”{o}dinger picture we have
\begin{aligned}{\lvert {\psi_s(t)} \rangle} &= U(t, t_0) {\lvert {\psi_s(t_0)} \rangle} \\ &=e^{-\frac{i}{\hbar} H_0(t - t_0)} U_I(t, t_0){\lvert {\psi_s(t_0)} \rangle}.\end{aligned}
With a definition of the interaction picture ket as
\begin{aligned}{\lvert {\psi_I} \rangle} = U_I(t, t_0) {\lvert {\psi_s(t_0)} \rangle} = U_I(t, t_0) {\lvert {\psi_H} \rangle},\end{aligned} \hspace{\stretch{1}}(2.24)
the Schr\”{o}dinger picture is then related to the interaction picture by
\begin{aligned}{\lvert {\psi_s(t)} \rangle} = e^{-\frac{i}{\hbar} H_0(t - t_0)} {\lvert {\psi_I} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.25)
Also, by multiplying 2.22 by our Schr\”{o}dinger ket, we remove the last vestiges of $U_I$ and $U$ from the dynamical equation for our time dependent interaction
\begin{aligned}i \hbar \frac{d{{}}}{dt} {\lvert {\psi_I} \rangle} = \bar{H}'(t) {\lvert {\psi_I} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.26)
## Interaction picture expectation.
Inverting 2.25, we can form an operator expectation, and relate it the interaction and Schr\”{o}dinger pictures
\begin{aligned}{\langle {\psi_s(t)} \rvert} O_s {\lvert {\psi_s(t)} \rangle} ={\langle {\psi_I} \rvert} e^{\frac{i}{\hbar} H_0(t - t_0)}O_se^{-\frac{i}{\hbar} H_0(t - t_0)}{\lvert {\psi_I} \rangle} .\end{aligned} \hspace{\stretch{1}}(2.27)
With a definition
\begin{aligned}O_I =e^{\frac{i}{\hbar} H_0(t - t_0)}O_se^{-\frac{i}{\hbar} H_0(t - t_0)},\end{aligned} \hspace{\stretch{1}}(2.28)
we have
\begin{aligned}{\langle {\psi_s(t)} \rvert} O_s {\lvert {\psi_s(t)} \rangle} ={\langle {\psi_I} \rvert} O_I{\lvert {\psi_I} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.29)
As before, the time evolution of our interaction picture operator, can be found by taking derivatives of 2.28, for which we find
\begin{aligned}i \hbar \frac{d{{O_I(t)}}}{dt} = \left[{O_I(t)},{H_0}\right]\end{aligned} \hspace{\stretch{1}}(2.30)
## Summarizing the interaction picture.
Given
\begin{aligned}H(t) = H_0 + H'(t),\end{aligned} \hspace{\stretch{1}}(2.31)
and initial time states
\begin{aligned}{\lvert {\psi_I(t_0)} \rangle} ={\lvert {\psi_s(t_0)} \rangle} = {\lvert {\psi_H} \rangle},\end{aligned} \hspace{\stretch{1}}(2.32)
we have
\begin{aligned}{\langle {\psi_s(t)} \rvert} O_s {\lvert {\psi_s(t)} \rangle} ={\langle {\psi_I} \rvert} O_I{\lvert {\psi_I} \rangle},\end{aligned} \hspace{\stretch{1}}(2.33)
where
\begin{aligned}{\lvert {\psi_I} \rangle} = U_I(t, t_0) {\lvert {\psi_s(t_0)} \rangle},\end{aligned} \hspace{\stretch{1}}(2.34)
and
\begin{aligned}i \hbar \frac{d{{}}}{dt} {\lvert {\psi_I} \rangle} = \bar{H}'(t) {\lvert {\psi_I} \rangle},\end{aligned} \hspace{\stretch{1}}(2.35)
or
\begin{aligned}i \hbar \frac{d{{}}}{dt} U_I(t, t_0) &= \bar{H}'(t) U_I(t, t_0) \\ U_I(t_0, t_0) &= I.\end{aligned} \hspace{\stretch{1}}(2.36)
Our interaction picture Hamiltonian is
\begin{aligned}\bar{H}'(t) =e^{\frac{i}{\hbar} H_0(t - t_0)} H'(t)) e^{-\frac{i}{\hbar} H_0(t - t_0)},\end{aligned} \hspace{\stretch{1}}(2.38)
and for Schr\”{o}dinger operators, independent of time, we have the dynamical equation
\begin{aligned}i \hbar \frac{d{{O_I(t)}}}{dt} = \left[{O_I(t)},{H_0}\right]\end{aligned} \hspace{\stretch{1}}(2.39)
# Justifying the Taylor expansion above (not class notes).
## Multivariable Taylor series
As outlined in section 2.8 ($8.10$) of [2], we want to derive the multi-variable Taylor expansion for a scalar valued function of some number of variables
\begin{aligned}f(\mathbf{u}) = f(u^1, u^2, \cdots),\end{aligned} \hspace{\stretch{1}}(3.40)
consider the displacement operation applied to the vector argument
\begin{aligned}f(\mathbf{a} + \mathbf{x}) = {\left.{{f(\mathbf{a} + t \mathbf{x})}}\right\vert}_{{t=1}}.\end{aligned} \hspace{\stretch{1}}(3.41)
We can Taylor expand a single variable function without any trouble, so introduce
\begin{aligned}g(t) = f(\mathbf{a} + t \mathbf{x}),\end{aligned} \hspace{\stretch{1}}(3.42)
where
\begin{aligned}g(1) = f(\mathbf{a} + \mathbf{x}).\end{aligned} \hspace{\stretch{1}}(3.43)
We have
\begin{aligned}g(t) = g(0) + t {\left.{{ \frac{\partial {g}}{\partial {t}} }}\right\vert}_{{t = 0}}+ \frac{t^2}{2!} {\left.{{ \frac{\partial {g}}{\partial {t}} }}\right\vert}_{{t = 0}}+ \cdots,\end{aligned} \hspace{\stretch{1}}(3.44)
so that
\begin{aligned}g(1) = g(0) + + {\left.{{ \frac{\partial {g}}{\partial {t}} }}\right\vert}_{{t = 0}}+ \frac{1}{2!} {\left.{{ \frac{\partial {g}}{\partial {t}} }}\right\vert}_{{t = 0}}+ \cdots.\end{aligned} \hspace{\stretch{1}}(3.45)
The multivariable Taylor series now becomes a plain old application of the chain rule, where we have to evaluate
\begin{aligned}\frac{dg}{dt} &= \frac{d{{}}}{dt} f(a^1 + t x^1, a^2 + t x^2, \cdots) \\ &= \sum_i \frac{\partial {}}{\partial {(a^i + t x^i)}} f(\mathbf{a} + t \mathbf{x}) \frac{\partial {a^i + t x^i}}{\partial {t}},\end{aligned}
so that
\begin{aligned}{\left.{{\frac{dg}{dt} }}\right\vert}_{{t=0}}= \sum_i x^i \left( {\left.{{ \frac{\partial {f}}{\partial {x^i}}}}\right\vert}_{{x^i = a^i}}\right).\end{aligned} \hspace{\stretch{1}}(3.46)
Assuming an Euclidean space we can write this in the notationally more pleasant fashion using a gradient operator for the space
\begin{aligned}{\left.{{\frac{dg}{dt} }}\right\vert}_{{t=0}} = {\left.{{\mathbf{x} \cdot \boldsymbol{\nabla}_{\mathbf{u}} f(\mathbf{u})}}\right\vert}_{{\mathbf{u} = \mathbf{a}}}.\end{aligned} \hspace{\stretch{1}}(3.47)
To handle the higher order terms, we repeat the chain rule application, yielding for example
\begin{aligned}{\left.{{\frac{d^2 f(\mathbf{a} + t \mathbf{x})}{dt^2} }}\right\vert}_{{t=0}} &={\left.{{\frac{d{{}}}{dt} \sum_i x^i \frac{\partial {f(\mathbf{a} + t \mathbf{x})}}{\partial {(a^i + t x^i)}} }}\right\vert}_{{t=0}}\\ &={\left.{{\sum_i x^i \frac{\partial {}}{\partial {(a^i + t x^i)}} \frac{d{{f(\mathbf{a} + t \mathbf{x})}}}{dt}}}\right\vert}_{{t=0}} \\ &={\left.{{(\mathbf{x} \cdot \boldsymbol{\nabla}_{\mathbf{u}})^2 f(\mathbf{u})}}\right\vert}_{{\mathbf{u} = \mathbf{a}}}.\end{aligned}
Thus the Taylor series associated with a vector displacement takes the tidy form
\begin{aligned}f(\mathbf{a} + \mathbf{x}) = \sum_{k=0}^\infty \frac{1}{{k!}} {\left.{{(\mathbf{x} \cdot \boldsymbol{\nabla}_{\mathbf{u}})^k f(\mathbf{u})}}\right\vert}_{{\mathbf{u} = \mathbf{a}}}.\end{aligned} \hspace{\stretch{1}}(3.48)
Even more fancy, we can form the operator equation
\begin{aligned}f(\mathbf{a} + \mathbf{x}) = {\left.{{e^{ \mathbf{x} \cdot \boldsymbol{\nabla}_{\mathbf{u}} } f(\mathbf{u})}}\right\vert}_{{\mathbf{u} = \mathbf{a}}}\end{aligned} \hspace{\stretch{1}}(3.49)
Here a dummy variable $\mathbf{u}$ has been retained as an instruction not to differentiate the $\mathbf{x}$ part of the directional derivative in any repeated applications of the $\mathbf{x} \cdot \boldsymbol{\nabla}$ operator.
That notational cludge can be removed by swapping $\mathbf{a}$ and $\mathbf{x}$
\begin{aligned}f(\mathbf{a} + \mathbf{x}) = \sum_{k=0}^\infty \frac{1}{{k!}} (\mathbf{a} \cdot \boldsymbol{\nabla})^k f(\mathbf{x})=e^{ \mathbf{a} \cdot \boldsymbol{\nabla} } f(\mathbf{x}),\end{aligned} \hspace{\stretch{1}}(3.50)
where $\boldsymbol{\nabla} = \boldsymbol{\nabla}_{\mathbf{x}} = ({\partial {}}/{\partial {x^1}}, {\partial {}}/{\partial {x^2}}, ...)$.
Having derived this (or for those with lesser degrees of amnesia, recall it), we can see that 2.2 was a direct application of this, retaining no second order or higher terms.
Our expression used in the interaction Hamiltonian discussion was
\begin{aligned}\frac{1}{{{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}}} \approx \frac{1}{{{\left\lvert{\mathbf{r}}\right\rvert}}} + \mathbf{R} \cdot {\left.{{\left(\frac{\partial {}}{\partial {\mathbf{R}}} \frac{1}{{{\left\lvert{ \mathbf{r} - \mathbf{R}}\right\rvert}}}\right)}}\right\vert}_{{\mathbf{R} = 0}}.\end{aligned} \hspace{\stretch{1}}(3.51)
which we can see has the same structure as above with some variable substitutions. Evaluating it we have
\begin{aligned}\frac{\partial {}}{\partial {\mathbf{R}}} \frac{1}{{{\left\lvert{ \mathbf{r} - \mathbf{R}}\right\rvert}}}&=\mathbf{e}_i \frac{\partial {}}{\partial {R^i}} ((x^j - R^j)^2)^{-1/2} \\ &=\mathbf{e}_i \left(-\frac{1}{{2}}\right) 2 (x^j - R^j) \frac{\partial {(x^j - R^j)}}{\partial {R^i}} \frac{1}{{{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}^3}} \\ &= \frac{\mathbf{r} - \mathbf{R}}{{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}^3} ,\end{aligned}
and at $\mathbf{R} = 0$ we have
\begin{aligned}\frac{1}{{{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}}} \approx \frac{1}{{{\left\lvert{\mathbf{r}}\right\rvert}}} + \mathbf{R} \cdot \frac{\mathbf{r}}{{\left\lvert{\mathbf{r}}\right\rvert}^3}.\end{aligned} \hspace{\stretch{1}}(3.52)
We see in this direction derivative produces the classical electric Coulomb field expression for an electrostatic distribution, once we take the $\mathbf{r}/{\left\lvert{\mathbf{r}}\right\rvert}^3$ and multiply it with the $- Z e$ factor.
## With algebra.
A different way to justify the expansion of 2.2 is to consider a Clifford algebra factorization (following notation from [3]) of the absolute vector difference, where $\mathbf{R}$ is considered small.
\begin{aligned}{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}&= \sqrt{ \left(\mathbf{r} - \mathbf{R}\right) \left(\mathbf{r} - \mathbf{R}\right) } \\ &= \sqrt{ \left\langle{{\mathbf{r} \left(1 - \frac{1}{\mathbf{r}} \mathbf{R}\right) \left(1 - \mathbf{R} \frac{1}{\mathbf{r}}\right) \mathbf{r}}}\right\rangle } \\ &= \sqrt{ \left\langle{{\mathbf{r}^2 \left(1 - \frac{1}{\mathbf{r}} \mathbf{R}\right) \left(1 - \mathbf{R} \frac{1}{\mathbf{r}}\right) }}\right\rangle } \\ &= {\left\lvert{\mathbf{r}}\right\rvert} \sqrt{ 1 - 2 \frac{1}{\mathbf{r}} \cdot \mathbf{R} + \left\langle{{\frac{1}{\mathbf{r}} \mathbf{R} \mathbf{R} \frac{1}{\mathbf{r}}}}\right\rangle} \\ &= {\left\lvert{\mathbf{r}}\right\rvert} \sqrt{ 1 - 2 \frac{1}{\mathbf{r}} \cdot \mathbf{R} + \frac{\mathbf{R}^2}{\mathbf{r}^2}}\end{aligned}
Neglecting the $\mathbf{R}^2$ term, we can then Taylor series expand this scalar expression
\begin{aligned}\frac{1}{{{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}}} \approx\frac{1}{{{\left\lvert{\mathbf{r}}\right\rvert}}} \left( 1 + \frac{1}{\mathbf{r}} \cdot \mathbf{R}\right) =\frac{1}{{{\left\lvert{\mathbf{r}}\right\rvert}}} + \frac{\hat{\mathbf{r}}}{\mathbf{r}^2} \cdot \mathbf{R}=\frac{1}{{{\left\lvert{\mathbf{r}}\right\rvert}}} + \frac{\mathbf{r}}{{\left\lvert{\mathbf{r}}\right\rvert}^3} \cdot \mathbf{R}.\end{aligned} \hspace{\stretch{1}}(3.53)
Observe this is what was found with the multivariable Taylor series expansion too.
# References
[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.
[2] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.
[3] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.
## PHY450HS1: Relativistic electrodynamics: some exam reflection.
Posted by peeterjoot on April 28, 2011
[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]
# Charged particle in a circle.
From the 2008 PHY353 exam, given a particle of charge $q$ moving in a circle of radius $a$ at constant angular frequency $\omega$.
\begin{itemize}
\item Find the Lienard-Wiechert potentials for points on the z-axis.
\item Find the electric and magnetic fields at the center.
\end{itemize}
When I tried this I did it for points not just on the z-axis. It turns out that we also got this question on the exam (but stated slightly differently). Since I’ll not get to see my exam solution again, let’s work through this at a leisurely rate, and see if things look right. The problem as stated in this old practice exam is easier since it doesn’t say to calculate the fields from the four potentials, so there was nothing preventing one from just grinding away and plugging stuff into the Lienard-Wiechert equations for the fields (as I did when I tried it for practice).
## The potentials.
Let’s set up our coordinate system in cylindrical coordinates. For the charged particle and the point that we measure the field, with $i = \mathbf{e}_1 \mathbf{e}_2$
\begin{aligned}\mathbf{x}(t) &= a \mathbf{e}_1 e^{i \omega t} \\ \mathbf{r} &= z \mathbf{e}_3 + \rho \mathbf{e}_1 e^{i \phi}\end{aligned} \hspace{\stretch{1}}(1.1)
Here I’m using the geometric product of vectors (if that’s unfamiliar then just substitute
\begin{aligned}\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\} \rightarrow \{\sigma_1, \sigma_2, \sigma_3\}\end{aligned} \hspace{\stretch{1}}(1.3)
We can do that since the Pauli matrices also have the same semantics (with a small difference since the geometric square of a unit vector is defined as the unit scalar, whereas the Pauli matrix square is the identity matrix). The semantics we require of this vector product are just $\mathbf{e}_\alpha^2 = 1$ and $\mathbf{e}_\alpha \mathbf{e}_\beta = - \mathbf{e}_\beta \mathbf{e}_\alpha$ for any $\alpha \ne \beta$.
I’ll also be loose with notation and use $\text{Real}(X) = \left\langle{{X}}\right\rangle$ to select the scalar part of a multivector (or with the Pauli matrices, the portion proportional to the identity matrix).
Our task is to compute the Lienard-Wiechert potentials. Those are
\begin{aligned}A^0 &= \frac{q}{R^{*}} \\ \mathbf{A} &= A^0 \frac{\mathbf{v}}{c},\end{aligned} \hspace{\stretch{1}}(1.4)
where
\begin{aligned}\mathbf{R} &= \mathbf{r} - \mathbf{x}(t_r) \\ R = {\left\lvert{\mathbf{R}}\right\rvert} &= c (t - t_r) \\ R^{*} &= R - \frac{\mathbf{v}}{c} \cdot \mathbf{R} \\ \mathbf{v} &= \frac{d\mathbf{x}}{dt_r}.\end{aligned} \hspace{\stretch{1}}(1.6)
We’ll need (eventually)
\begin{aligned}\mathbf{v} &= a \omega \mathbf{e}_2 e^{i \omega t_r} = a \omega ( -\sin \omega t_r, \cos\omega t_r, 0) \\ \dot{\mathbf{v}} &= -a \omega^2 \mathbf{e}_1 e^{i \omega t_r} = -a \omega^2 (\cos\omega t_r, \sin\omega t_r, 0)\end{aligned} \hspace{\stretch{1}}(1.10)
and also need our retarded distance vector
\begin{aligned}\mathbf{R} = z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a e^{i \omega t_r} ),\end{aligned} \hspace{\stretch{1}}(1.12)
From this we have
\begin{aligned}R^2 &= z^2 + {\left\lvert{\mathbf{e}_1 (\rho e^{i \phi} - a e^{i \omega t_r} )}\right\rvert}^2 \\ &= z^2 + \rho^2 + a^2 - 2 \rho a (\mathbf{e}_1 \rho e^{i \phi}) \cdot (\mathbf{e}_1 e^{i \omega t_r}) \\ &= z^2 + \rho^2 + a^2 - 2 \rho a \text{Real}( e^{ i(\phi - \omega t_r) } ) \\ &= z^2 + \rho^2 + a^2 - 2 \rho a \cos(\phi - \omega t_r)\end{aligned}
So
\begin{aligned}R = \sqrt{z^2 + \rho^2 + a^2 - 2 \rho a \cos( \phi - \omega t_r ) }.\end{aligned} \hspace{\stretch{1}}(1.13)
Next we need
\begin{aligned}\mathbf{R} \cdot \mathbf{v}/c&= (z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a e^{i \omega t_r} )) \cdot \left(a \frac{\omega}{c} \mathbf{e}_2 e^{i \omega t_r} \right) \\ &=a \frac{\omega }{c}\text{Real}(i (\rho e^{-i \phi} - a e^{-i \omega t_r} ) e^{i \omega t_r} ) \\ &=a \frac{\omega }{c}\rho \text{Real}( i e^{-i \phi + i \omega t_r} ) \\ &=a \frac{\omega }{c}\rho \sin(\phi - \omega t_r)\end{aligned}
So we have
\begin{aligned}R^{*} = \sqrt{z^2 + \rho^2 + a^2 - 2 \rho a \cos( \phi - \omega t_r ) }-a \frac{\omega }{c} \rho \sin(\phi - \omega t_r)\end{aligned} \hspace{\stretch{1}}(1.14)
Writing $k = \omega/c$, and having a peek back at 1.4, our potentials are now solved for
\begin{aligned}\boxed{\begin{aligned}A^0 &= \frac{q}{\sqrt{z^2 + \rho^2 + a^2 - 2 \rho a \cos( \phi - k c t_r ) }} \\ \mathbf{A} &= A^0 a k ( -\sin k c t_r, \cos k c t_r, 0).\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.24)
The caveat is that $t_r$ is only specified implicitly, according to
\begin{aligned}\boxed{c t_r = c t - \sqrt{z^2 + \rho^2 + a^2 - 2 \rho a \cos( \phi - k c t_r ) }.}\end{aligned} \hspace{\stretch{1}}(1.16)
There doesn’t appear to be much hope of solving for $t_r$ explicitly in closed form.
## General fields for this system.
With
\begin{aligned}\mathbf{R}^{*} = \mathbf{R} - \frac{\mathbf{v}}{c} R,\end{aligned} \hspace{\stretch{1}}(1.17)
the fields are
\begin{aligned}\boxed{\begin{aligned}\mathbf{E} &= q (1 - \mathbf{v}^2/c^2) \frac{\mathbf{R}^{*}}{{R^{*}}^3} + \frac{q}{{R^{*}}^3} \mathbf{R} \times (\mathbf{R}^{*} \times \dot{\mathbf{v}}/c^2) \\ \mathbf{B} &= \frac{\mathbf{R}}{R} \times \mathbf{E}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.18)
In there we have
\begin{aligned}1 - \mathbf{v}^2/c^2 = 1 - a^2 \frac{\omega^2}{c^2} = 1 - a^2 k^2\end{aligned} \hspace{\stretch{1}}(1.19)
and
\begin{aligned}\mathbf{R}^{*} &= z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a e^{i k c t_r} )-a k \mathbf{e}_2 e^{i k c t_r} R \\ &= z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a (1 - k R i) e^{i k c t_r} )\end{aligned}
Writing this out in coordinates isn’t particularly illuminating, but can be done for completeness without too much trouble
\begin{aligned}\mathbf{R}^{*} = ( \rho \cos\phi - a \cos t_r + a k R \sin t_r, \rho \sin\phi - a \sin t_r - a k R \cos t_r, z )\end{aligned} \hspace{\stretch{1}}(1.20)
In one sense the problem could be considered solved, since we have all the pieces of the puzzle. The outstanding question is whether or not the resulting mess can be simplified at all. Let’s see if the cross product reduces at all. Using
\begin{aligned}\mathbf{R} \times (\mathbf{R}^{*} \times \dot{\mathbf{v}}/c^2) =\mathbf{R}^{*} (\mathbf{R} \cdot \dot{\mathbf{v}}/c^2) - \frac{\dot{\mathbf{v}}}{c^2}(\mathbf{R} \cdot \mathbf{R}^{*})\end{aligned} \hspace{\stretch{1}}(1.21)
Perhaps one or more of these dot products can be simplified? One of them does reduce nicely
\begin{aligned}\mathbf{R}^{*} \cdot \mathbf{R} &= ( \mathbf{R} - R \mathbf{v}/c ) \cdot \mathbf{R} \\ &= R^2 - (\mathbf{R} \cdot \mathbf{v}/c) R \\ &= R^2 - R a k \rho \sin(\phi - k c t_r) \\ &= R(R - a k \rho \sin(\phi - k c t_r))\end{aligned}
\begin{aligned}\mathbf{R} \cdot \dot{\mathbf{v}}/c^2&=\Bigl(z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a e^{i \omega t_r} ) \Bigr) \cdot(-a k^2 \mathbf{e}_1 e^{i \omega t_r} ) \\ &=- a k^2 \left\langle{{\mathbf{e}_1 (\rho e^{i \phi} - a e^{i \omega t_r} ) \mathbf{e}_1 e^{i \omega t_r} ) }}\right\rangle \\ &=- a k^2 \left\langle{{(\rho e^{i \phi} - a e^{i \omega t_r} ) e^{-i \omega t_r} ) }}\right\rangle \\ &=- a k^2 \left\langle{{\rho e^{i \phi - i \omega t_r} - a }}\right\rangle \\ &=- a k^2 ( \rho \cos(\phi - k c t_r) - a )\end{aligned}
Putting this cross product back together we have
\begin{aligned}\mathbf{R} \times (\mathbf{R}^{*} \times \dot{\mathbf{v}}/c^2)&=a k^2 ( a -\rho \cos(\phi - k c t_r) ) \mathbf{R}^{*} +a k^2 \mathbf{e}_1 e^{i k c t_r} R(R - a k \rho \sin(\phi - k c t_r)) \\ &=a k^2 ( a -\rho \cos(\phi - k c t_r) ) \Bigl(z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a (1 - k R i) e^{i k c t_r} )\Bigr) \\ &\qquad +a k^2 R \mathbf{e}_1 e^{i k c t_r} (R - a k \rho \sin(\phi - k c t_r)) \end{aligned}
Writing
\begin{aligned}\phi_r = \phi - k c t_r,\end{aligned} \hspace{\stretch{1}}(1.22)
this can be grouped into similar terms
\begin{aligned}\begin{aligned}\mathbf{R} \times (\mathbf{R}^{*} \times \dot{\mathbf{v}}/c^2)&=a k^2 (a - \rho \cos\phi_r) z \mathbf{e}_3 \\ &+ a k^2 \mathbf{e}_1(a - \rho \cos\phi_r) \rho e^{i\phi} \\ &+ a k^2 \mathbf{e}_1\left(-a (a - \rho \cos\phi_r) (1 - k R i)+ R(R - a k \rho \sin \phi_r)\right) e^{i k c t_r}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.23)
The electric field pieces can now be collected. Not expanding out the $R^{*}$ from 1.14, this is
\begin{aligned}\begin{aligned}\mathbf{E} &= \frac{q}{(R^{*})^3} z \mathbf{e}_3\Bigl( 1 - a \rho k^2 \cos\phi_r \Bigr) \\ &+\frac{q}{(R^{*})^3} \rho\mathbf{e}_1 \Bigl(1 - a \rho k^2 \cos\phi_r \Bigr) e^{i\phi} \\ &+\frac{q}{(R^{*})^3} a \mathbf{e}_1\left(-\Bigl( 1 + a k^2 (a - \rho \cos\phi_r) \Bigr) (1 - k R i)(1 - a^2 k^2)+ k^2 R(R - a k \rho \sin \phi_r)\right) e^{i k c t_r}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.24)
Along the z-axis where $\rho = 0$ what do we have?
\begin{aligned}R = \sqrt{z^2 + a^2 } \end{aligned} \hspace{\stretch{1}}(1.25)
\begin{aligned}A^0 = \frac{q}{R} \end{aligned} \hspace{\stretch{1}}(1.26)
\begin{aligned}\mathbf{A} = A^0 a k \mathbf{e}_2 e^{i k c t_r } \end{aligned} \hspace{\stretch{1}}(1.27)
\begin{aligned}c t_r = c t - \sqrt{z^2 + a^2 } \end{aligned} \hspace{\stretch{1}}(1.28)
\begin{aligned}\begin{aligned}\mathbf{E} &= \frac{q}{R^3} z \mathbf{e}_3 \\ &+\frac{q}{R^3} a \mathbf{e}_1\left(-( 1 - a^4 k^4 ) (1 - k R i)+ k^2 R^2 \right) e^{i k c t_r} \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.29)
\begin{aligned}\mathbf{B} = \frac{ z \mathbf{e}_3 - a \mathbf{e}_1 e^{i k c t_r}}{R} \times \mathbf{E}\end{aligned} \hspace{\stretch{1}}(1.30)
The magnetic term here looks like it can be reduced a bit.
## An approximation near the center.
Unlike the old exam I did, where it didn’t specify that the potentials had to be used to calculate the fields, and the problem was reduced to one of algebraic manipulation, our exam explicitly asked for the potentials to be used to calculate the fields.
There was also the restriction to compute them near the center. Setting $\rho = 0$ so that we are looking only near the z-axis, we have
\begin{aligned}A^0 &= \frac{q}{\sqrt{z^2 + a^2}} \\ \mathbf{A} &= \frac{q a k \mathbf{e}_2 e^{i k c t_r} }{\sqrt{z^2 + a^2}} = \frac{q a k (-\sin k c t_r, \cos k c t_r, 0)}{\sqrt{z^2 + a^2}} \\ t_r &= t - R/c = t - \sqrt{z^2 + a^2}/c\end{aligned} \hspace{\stretch{1}}(1.31)
Now we are set to calculate the electric and magnetic fields directly from these. Observe that we have a spatial dependence in due to the $t_r$ quantities and that will have an effect when we operate with the gradient.
In the exam I’d asked Simon (our TA) if this question was asking for the fields at the origin (ie: in the plane of the charge’s motion in the center) or along the z-axis. He said in the plane. That would simplify things, but perhaps too much since $A^0$ becomes constant (in my exam attempt I somehow fudged this to get what I wanted for the $v = 0$ case, but that must have been wrong, and was the result of rushed work).
Let’s now proceed with the field calculation from these potentials
\begin{aligned}\mathbf{E} &= - \boldsymbol{\nabla} A^0 - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \\ \mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(1.34)
For the electric field we need
\begin{aligned}\boldsymbol{\nabla} A^0 &= q \mathbf{e}_3 \partial_z (z^2 + a^2)^{-1/2} \\ &= -q \mathbf{e}_3 \frac{z}{(\sqrt{z^2 + a^2})^3},\end{aligned}
and
\begin{aligned}\frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} =\frac{q a k^2 \mathbf{e}_2 \mathbf{e}_1 \mathbf{e}_2 e^{i k c t_r} }{\sqrt{z^2 + a^2}}.\end{aligned} \hspace{\stretch{1}}(1.36)
Putting these together, our electric field near the z-axis is
\begin{aligned}\mathbf{E} = q \mathbf{e}_3 \frac{z}{(\sqrt{z^2 + a^2})^3}+\frac{q a k^2 \mathbf{e}_1 e^{i k c t_r} }{\sqrt{z^2 + a^2}}.\end{aligned} \hspace{\stretch{1}}(1.37)
(another mistake I made on the exam, since I somehow fooled myself into forcing what I knew had to be in the gradient term, despite having essentially a constant scalar potential (having taken $z = 0$)).
What do we get for the magnetic field. In that case we have
\begin{aligned}\boldsymbol{\nabla} \times \mathbf{A}(z)&=\mathbf{e}_\alpha \times \partial_\alpha \mathbf{A} \\ &=\mathbf{e}_3 \times \partial_z \frac{q a k \mathbf{e}_2 e^{i k c t_r} }{\sqrt{z^2 + a^2}} \\ &=\mathbf{e}_3 \times (\mathbf{e}_2 e^{i k c t_r} ) q a k \frac{\partial {}}{\partial {z}} \frac{1}{{\sqrt{z^2 + a^2}}} +q a k \frac{1}{{\sqrt{z^2 + a^2}}} \mathbf{e}_3 \times (\mathbf{e}_2 \partial_z e^{i k c t_r} ) \\ &=-\mathbf{e}_3 \times (\mathbf{e}_2 e^{i k c t_r} ) q a k \frac{z}{(\sqrt{z^2 + a^2})^3} +q a k \frac{1}{{\sqrt{z^2 + a^2}}} \mathbf{e}_3 \times \left( \mathbf{e}_2 \mathbf{e}_1 \mathbf{e}_2 k c e^{i k c t_r} \partial_z ( t - \sqrt{z^a + a^2}/c ) \right) \\ &=-\mathbf{e}_3 \times (\mathbf{e}_2 e^{i k c t_r} ) q a k \frac{z}{(\sqrt{z^2 + a^2})^3} -q a k^2 \frac{z}{z^2 + a^2} \mathbf{e}_3 \times \left( \mathbf{e}_1 k e^{i k c t_r} \right) \\ &=-\frac{q a k z \mathbf{e}_3}{z^2 + a^2} \times \left( \frac{ \mathbf{e}_2 e^{i k c t_r} }{\sqrt{z^2 + a^2}} + k \mathbf{e}_1 e^{i k c t_r} \right)\end{aligned}
For the direction vectors in the cross products above we have
\begin{aligned}\mathbf{e}_3 \times (\mathbf{e}_2 e^{i \mu})&=\mathbf{e}_3 \times (\mathbf{e}_2 \cos\mu - \mathbf{e}_1 \sin\mu) \\ &=-\mathbf{e}_1 \cos\mu - \mathbf{e}_2 \sin\mu \\ &=-\mathbf{e}_1 e^{i \mu}\end{aligned}
and
\begin{aligned}\mathbf{e}_3 \times (\mathbf{e}_1 e^{i \mu})&=\mathbf{e}_3 \times (\mathbf{e}_1 \cos\mu + \mathbf{e}_2 \sin\mu) \\ &=\mathbf{e}_2 \cos\mu - \mathbf{e}_1 \sin\mu \\ &=\mathbf{e}_2 e^{i \mu}\end{aligned}
Putting everything, and summarizing results for the fields, we have
\begin{aligned}\mathbf{E} &= q \mathbf{e}_3 \frac{z}{(\sqrt{z^2 + a^2})^3}+\frac{q a k^2 \mathbf{e}_1 e^{i \omega t_r} }{\sqrt{z^2 + a^2}} \\ \mathbf{B} &= \frac{q a k z}{ z^2 + a^2} \left( \frac{\mathbf{e}_1}{\sqrt{z^2 + a^2}} - k \mathbf{e}_2 \right) e^{i \omega t_r}\end{aligned} \hspace{\stretch{1}}(1.38)
The electric field expression above compares well to 1.29. We have the Coulomb term and the radiation term. It is harder to compare the magnetic field to the exact result 1.30 since I did not expand that out.
FIXME: A question to consider. If all this worked should we not also get
\begin{aligned}\mathbf{B} \stackrel{?}{=}\frac{z \mathbf{e}_3 - \mathbf{e}_1 a e^{i \omega t_r}}{\sqrt{z^2 + a^2}} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.40)
However, if I do this check I get
\begin{aligned}\mathbf{B} =\frac{q a z}{z^2 + a^2} \left( \frac{1}{{z^2 + a^2}} + k^2 \right) \mathbf{e}_2 e^{i \omega t_r}.\end{aligned} \hspace{\stretch{1}}(1.41)
# Collision of photon and electron.
I made a dumb error on the exam on this one. I setup the four momentum conservation statement, but then didn’t multiply out the cross terms properly. This led me to incorrectly assume that I had to try doing this the hard way (something akin to what I did on the midterm). Simon later told us in the tutorial the simple way, and that’s all we needed here too. Here’s the setup.
An electron at rest initially has four momentum
\begin{aligned}(m c, 0)\end{aligned} \hspace{\stretch{1}}(2.42)
where the incoming photon has four momentum
\begin{aligned}\left(\hbar \frac{\omega}{c}, \hbar \mathbf{k} \right)\end{aligned} \hspace{\stretch{1}}(2.43)
After the collision our electron has some velocity so its four momentum becomes (say)
\begin{aligned}\gamma (m c, m \mathbf{v}),\end{aligned} \hspace{\stretch{1}}(2.44)
and our new photon, going off on an angle $\theta$ relative to $\mathbf{k}$ has four momentum
\begin{aligned}\left(\hbar \frac{\omega'}{c}, \hbar \mathbf{k}' \right)\end{aligned} \hspace{\stretch{1}}(2.45)
Our conservation relationship is thus
\begin{aligned}(m c, 0) + \left(\hbar \frac{\omega}{c}, \hbar \mathbf{k} \right)=\gamma (m c, m \mathbf{v})+\left(\hbar \frac{\omega'}{c}, \hbar \mathbf{k}' \right)\end{aligned} \hspace{\stretch{1}}(2.46)
I squared both sides, but dropped my cross terms, which was just plain wrong, and costly for both time and effort on the exam. What I should have done was just
\begin{aligned}\gamma (m c, m \mathbf{v}) =(m c, 0) + \left(\hbar \frac{\omega}{c}, \hbar \mathbf{k} \right)-\left(\hbar \frac{\omega'}{c}, \hbar \mathbf{k}' \right),\end{aligned} \hspace{\stretch{1}}(2.47)
and then square this (really making contractions of the form $p_i p^i$). That gives (and this time keeping my cross terms)
\begin{aligned}(\gamma (m c, m \mathbf{v}) )^2 &= \gamma^2 m^2 (c^2 - \mathbf{v}^2) \\ &= m^2 c^2 \\ &=m^2 c^2 + 0 + 0+ 2 (m c, 0) \cdot \left(\hbar \frac{\omega}{c}, \hbar \mathbf{k} \right)- 2 (m c, 0) \cdot \left(\hbar \frac{\omega'}{c}, \hbar \mathbf{k}' \right)- 2 \cdot \left(\hbar \frac{\omega}{c}, \hbar \mathbf{k} \right)\cdot \left(\hbar \frac{\omega'}{c}, \hbar \mathbf{k}' \right) \\ &=m^2 c^2 + 2 m c \hbar \frac{\omega}{c} - 2 m c \hbar \frac{\omega'}{c}- 2\hbar^2 \left(\frac{\omega}{c} \frac{\omega'}{c}- \mathbf{k} \cdot \mathbf{k}'\right) \\ &=m^2 c^2 + 2 m c \hbar \frac{\omega}{c} - 2 m c \hbar \frac{\omega'}{c}- 2\hbar^2 \frac{\omega}{c} \frac{\omega'}{c} (1 - \cos\theta)\end{aligned}
Rearranging a bit we have
\begin{aligned}\omega' \left( m + \frac{\hbar \omega}{c^2} ( 1 - \cos\theta ) \right) = m \omega,\end{aligned} \hspace{\stretch{1}}(2.48)
or
\begin{aligned}\omega' = \frac{\omega}{1 + \frac{\hbar \omega}{m c^2} ( 1 - \cos\theta ) }\end{aligned} \hspace{\stretch{1}}(2.49)
# Pion decay.
The problem above is very much like a midterm problem we had, so there was no justifiable excuse for messing up on it. That midterm problem was to consider the split of a pion at rest into a neutrino (massless) and a muon, and to calculate the energy of the muon. That one also follows the same pattern, a calculation of four momentum conservation, say
\begin{aligned}(m_\pi c, 0) = \hbar \frac{\omega}{c}(1, \hat{\mathbf{k}}) + ( \mathcal{E}_\mu/c, \mathbf{p}_\mu ).\end{aligned} \hspace{\stretch{1}}(3.50)
Here $\omega$ is the frequency of the massless neutrino. The massless nature is encoded by a four momentum that squares to zero, which follows from $(1, \hat{\mathbf{k}}) \cdot (1, \hat{\mathbf{k}}) = 1^2 - \hat{\mathbf{k}} \cdot \hat{\mathbf{k}} = 0$.
When I did this problem on the midterm, I perversely put in a scattering angle, instead of recognizing that the particles must scatter at 180 degree directions since spatial momentum components must also be preserved. This and the combination of trying to work in spatial quantities led to a mess and I didn’t get the end result in anything that could be considered tidy.
The simple way to do this is to just rearrange to put the null vector on one side, and then square. This gives us
\begin{aligned}0 &=\left(\hbar \frac{\omega}{c}(1, \hat{\mathbf{k}}) \right) \cdot\left(\hbar \frac{\omega}{c}(1, \hat{\mathbf{k}}) \right) \\ &=\left( (m_\pi c, 0) - ( \mathcal{E}_\mu/c, \mathbf{p}_\mu ) \right) \cdot \left( (m_\pi c, 0) - ( \mathcal{E}_\mu/c, \mathbf{p}_\mu ) \right) \\ &={m_\pi}^2 c^2 + {m_\nu}^2 c^2 - 2 (m_\pi c, 0) \cdot ( \mathcal{E}_\mu/c, \mathbf{p}_\mu ) \\ &={m_\pi}^2 c^2 + {m_\nu}^2 c^2 - 2 m_\pi \mathcal{E}_\mu\end{aligned}
A final re-arrangement gives us the muon energy
\begin{aligned}\mathcal{E}_\mu = \frac{1}{{2}} \frac{ {m_\pi}^2 + {m_\nu}^2 }{m_\pi} c^2\end{aligned} \hspace{\stretch{1}}(3.51)
## Fourier transform solutions and associated energy and momentum for the homogeneous Maxwell equation. (rework once more)
Posted by peeterjoot on December 29, 2009
[Click here for a PDF of this post with nicer formatting]. Note that this PDF file is formatted in a wide-for-screen layout that is probably not good for printing.
These notes build on and replace those formerly posted in Energy and momentum for assumed Fourier transform solutions to the homogeneous Maxwell equation.
# Motivation and notation.
In Electrodynamic field energy for vacuum (reworked) [1], building on Energy and momentum for Complex electric and magnetic field phasors [2], a derivation for the energy and momentum density was derived for an assumed Fourier series solution to the homogeneous Maxwell’s equation. Here we move to the continuous case examining Fourier transform solutions and the associated energy and momentum density.
A complex (phasor) representation is implied, so taking real parts when all is said and done is required of the fields. For the energy momentum tensor the Geometric Algebra form, modified for complex fields, is used
\begin{aligned}T(a) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \hspace{\stretch{1}}(1.1)
The assumed four vector potential will be written
\begin{aligned}A(\mathbf{x}, t) = A^\mu(\mathbf{x}, t) \gamma_\mu = \frac{1}{{(\sqrt{2 \pi})^3}} \int A(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(1.2)
Subject to the requirement that $A$ is a solution of Maxwell’s equation
\begin{aligned}\nabla (\nabla \wedge A) = 0.\end{aligned} \hspace{\stretch{1}}(1.3)
To avoid latex hell, no special notation will be used for the Fourier coefficients,
\begin{aligned}A(\mathbf{k}, t) = \frac{1}{{(\sqrt{2 \pi})^3}} \int A(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{x}.\end{aligned} \hspace{\stretch{1}}(1.4)
When convenient and unambiguous, this $(\mathbf{k},t)$ dependence will be implied.
Having picked a time and space representation for the field, it will be natural to express both the four potential and the gradient as scalar plus spatial vector, instead of using the Dirac basis. For the gradient this is
\begin{aligned}\nabla &= \gamma^\mu \partial_\mu = (\partial_0 - \boldsymbol{\nabla}) \gamma_0 = \gamma_0 (\partial_0 + \boldsymbol{\nabla}),\end{aligned} \hspace{\stretch{1}}(1.5)
and for the four potential (or the Fourier transform functions), this is
\begin{aligned}A &= \gamma_\mu A^\mu = (\phi + \mathbf{A}) \gamma_0 = \gamma_0 (\phi - \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(1.6)
# Setup
The field bivector $F = \nabla \wedge A$ is required for the energy momentum tensor. This is
\begin{aligned}\nabla \wedge A&= \frac{1}{{2}}\left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \\ &= \frac{1}{{2}}\left( (\stackrel{ \rightarrow }{\partial}_0 - \stackrel{ \rightarrow }{\boldsymbol{\nabla}}) \gamma_0 \gamma_0 (\phi - \mathbf{A})-(\phi + \mathbf{A}) \gamma_0 \gamma_0 (\stackrel{ \leftarrow }{\partial}_0 + \stackrel{ \leftarrow }{\boldsymbol{\nabla}})\right) \\ &= -\boldsymbol{\nabla} \phi -\partial_0 \mathbf{A} + \frac{1}{{2}}(\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{A} - \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}})\end{aligned}
This last term is a spatial curl and the field is then
\begin{aligned}F = -\boldsymbol{\nabla} \phi -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A}\end{aligned} \hspace{\stretch{1}}(2.7)
Applied to the Fourier representation this is
\begin{aligned}F =\frac{1}{{(\sqrt{2 \pi})^3}} \int\left(- \frac{1}{c} \dot{\mathbf{A}}- i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(2.8)
It is only the real parts of this that we are actually interested in, unless physical meaning can be assigned to the complete complex vector field.
# Constraints supplied by Maxwell’s equation.
A Fourier transform solution of Maxwell’s vacuum equation $\nabla F = 0$ has been assumed. Having expressed the Faraday bivector in terms of spatial vector quantities, it is more convenient to do this back substitution into after pre-multiplying Maxwell’s equation by $\gamma_0$, namely
\begin{aligned}0&= \gamma_0 \nabla F \\ &= (\partial_0 + \boldsymbol{\nabla}) F.\end{aligned} \hspace{\stretch{1}}(3.9)
Applied to the spatially decomposed field as specified in (2.7), this is
\begin{aligned}0&=-\partial_0 \boldsymbol{\nabla} \phi-\partial_{00} \mathbf{A}+ \partial_0 \boldsymbol{\nabla} \wedge \mathbf{A}-\boldsymbol{\nabla}^2 \phi- \boldsymbol{\nabla} \partial_0 \mathbf{A}+ \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &=- \partial_0 \boldsymbol{\nabla} \phi - \boldsymbol{\nabla}^2 \phi- \partial_{00} \mathbf{A}- \boldsymbol{\nabla} \cdot \partial_0 \mathbf{A}+ \boldsymbol{\nabla}^2 \mathbf{A} - \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} ) \\ \end{aligned}
All grades of this equation must simultaneously equal zero, and the bivector grades have canceled (assuming commuting space and time partials), leaving two equations of constraint for the system
\begin{aligned}0 &=\boldsymbol{\nabla}^2 \phi + \boldsymbol{\nabla} \cdot \partial_0 \mathbf{A}\end{aligned} \hspace{\stretch{1}}(3.11)
\begin{aligned}0 &=\partial_{00} \mathbf{A} - \boldsymbol{\nabla}^2 \mathbf{A}+ \boldsymbol{\nabla} \partial_0 \phi + \boldsymbol{\nabla} ( \boldsymbol{\nabla} \cdot \mathbf{A} )\end{aligned} \hspace{\stretch{1}}(3.12)
It is immediately evident that a gauge transformation could be immediately helpful to simplify things. In [3] the gauge choice $\boldsymbol{\nabla} \cdot \mathbf{A} = 0$ is used. From (3.11) this implies that $\boldsymbol{\nabla}^2 \phi = 0$. Bohm argues that for this current and charge free case this implies $\phi = 0$, but he also has a periodicity constraint. Without a periodicity constraint it is easy to manufacture non-zero counterexamples. One is a linear function in the space and time coordinates
\begin{aligned}\phi = p x + q y + r z + s t\end{aligned} \hspace{\stretch{1}}(3.13)
This is a valid scalar potential provided that the wave equation for the vector potential is also a solution. We can however, force $\phi = 0$ by making the transformation $A^\mu \rightarrow A^\mu + \partial^\mu \psi$, which in non-covariant notation is
\begin{aligned}\phi &\rightarrow \phi + \frac{1}{c} \partial_t \psi \\ \mathbf{A} &\rightarrow \phi - \boldsymbol{\nabla} \psi\end{aligned} \hspace{\stretch{1}}(3.14)
If the transformed field $\phi' = \phi + \partial_t \psi/c$ can be forced to zero, then the complexity of the associated Maxwell equations are reduced. In particular, antidifferentiation of $\phi = -(1/c) \partial_t \psi$, yields
\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x}, 0) - c \int_{\tau=0}^t \phi(\mathbf{x}, \tau) d\tau.\end{aligned} \hspace{\stretch{1}}(3.16)
Dropping primes, the transformed Maxwell equations now take the form
\begin{aligned}0 &= \partial_t( \boldsymbol{\nabla} \cdot \mathbf{A} )\end{aligned} \hspace{\stretch{1}}(3.17)
\begin{aligned}0 &=\partial_{00} \mathbf{A} - \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} ).\end{aligned} \hspace{\stretch{1}}(3.18)
There are two classes of solutions that stand out for these equations. If the vector potential is constant in time $\mathbf{A}(\mathbf{x},t) = \mathbf{A}(\mathbf{x})$, Maxwell’s equations are reduced to the single equation
\begin{aligned}0&= - \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} ).\end{aligned} \hspace{\stretch{1}}(3.19)
Observe that a gradient can be factored out of this equation
\begin{aligned}- \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} )&=\boldsymbol{\nabla} (-\boldsymbol{\nabla} \mathbf{A} + \boldsymbol{\nabla} \cdot \mathbf{A} ) \\ &=-\boldsymbol{\nabla} (\boldsymbol{\nabla} \wedge \mathbf{A}).\end{aligned}
The solutions are then those $\mathbf{A}$s that satisfy both
\begin{aligned}0 &= \partial_t \mathbf{A} \\ 0 &= \boldsymbol{\nabla} (\boldsymbol{\nabla} \wedge \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(3.20)
In particular any non-time dependent potential $\mathbf{A}$ with constant curl provides a solution to Maxwell’s equations. There may be other solutions to (3.19) too that are more general. Returning to (3.17) a second way to satisfy these equations stands out. Instead of requiring of $\mathbf{A}$ constant curl, constant divergence with respect to the time partial eliminates (3.17). The simplest resulting equations are those for which the divergence is a constant in time and space (such as zero). The solution set are then spanned by the vectors $\mathbf{A}$ for which
\begin{aligned}\text{constant} &= \boldsymbol{\nabla} \cdot \mathbf{A} \end{aligned} \hspace{\stretch{1}}(3.22)
\begin{aligned}0 &= \frac{1}{{c^2}} \partial_{tt} \mathbf{A} - \boldsymbol{\nabla}^2 \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(3.23)
Any $\mathbf{A}$ that both has constant divergence and satisfies the wave equation will via (2.7) then produce a solution to Maxwell’s equation.
# Maxwell equation constraints applied to the assumed Fourier solutions.
Let’s consider Maxwell’s equations in all three forms, (3.11), (3.20), and (3.22) and apply these constraints to the assumed Fourier solution.
In all cases the starting point is a pair of Fourier transform relationships, where the Fourier transforms are the functions to be determined
\begin{aligned}\phi(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \phi(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \end{aligned} \hspace{\stretch{1}}(4.24)
\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \mathbf{A}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \end{aligned} \hspace{\stretch{1}}(4.25)
## Case I. Constant time vector potential. Scalar potential eliminated by gauge transformation.
From (4.24) we require
\begin{aligned}0 = (2 \pi)^{-3/2} \int \partial_t \mathbf{A}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.26)
So the Fourier transform also cannot have any time dependence, and we have
\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \mathbf{A}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \end{aligned} \hspace{\stretch{1}}(4.27)
What is the curl of this? Temporarily falling back to coordinates is easiest for this calculation
\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{A}(\mathbf{k}) e^{i\mathbf{k} \cdot \mathbf{x}}&=\sigma_m \partial_m \wedge \sigma_n A^n(\mathbf{k}) e^{i \mathbf{x} \cdot \mathbf{x}} \\ &=\sigma_m \wedge \sigma_n A^n(\mathbf{k}) i k^m e^{i \mathbf{x} \cdot \mathbf{x}} \\ &=i\mathbf{k} \wedge \mathbf{A}(\mathbf{k}) e^{i \mathbf{x} \cdot \mathbf{x}} \\ \end{aligned}
This gives
\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.28)
We want to equate the divergence of this to zero. Neglecting the integral and constant factor this requires
\begin{aligned}0 &= \boldsymbol{\nabla} \cdot \left( i \mathbf{k} \wedge \mathbf{A} e^{i\mathbf{k} \cdot \mathbf{x}} \right) \\ &= {\left\langle{{ \sigma_m \partial_m i (\mathbf{k} \wedge \mathbf{A}) e^{i\mathbf{k} \cdot \mathbf{x}} }}\right\rangle}_{1} \\ &= -{\left\langle{{ \sigma_m (\mathbf{k} \wedge \mathbf{A}) k^m e^{i\mathbf{k} \cdot \mathbf{x}} }}\right\rangle}_{1} \\ &= -\mathbf{k} \cdot (\mathbf{k} \wedge \mathbf{A}) e^{i\mathbf{k} \cdot \mathbf{x}} \\ \end{aligned}
Requiring that the plane spanned by $\mathbf{k}$ and $\mathbf{A}(\mathbf{k})$ be perpendicular to $\mathbf{k}$ implies that $\mathbf{A} \propto \mathbf{k}$. The solution set is then completely described by functions of the form
\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \mathbf{k} \psi(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k},\end{aligned} \hspace{\stretch{1}}(4.29)
where $\psi(\mathbf{k})$ is an arbitrary scalar valued function. This is however, an extremely uninteresting solution since the curl is uniformly zero
\begin{aligned}F &= \boldsymbol{\nabla} \wedge \mathbf{A} \\ &= (2 \pi)^{-3/2} \int (i \mathbf{k}) \wedge \mathbf{k} \psi(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned}
Since $\mathbf{k} \wedge \mathbf{k} = 0$, when all is said and done the $\phi = 0$, $\partial_t \mathbf{A} = 0$ case appears to have no non-trivial (zero) solutions. Moving on, …
## Case II. Constant vector potential divergence. Scalar potential eliminated by gauge transformation.
Next in the order of complexity is consideration of the case (3.22). Here we also have $\phi = 0$, eliminated by gauge transformation, and are looking for solutions with the constraint
\begin{aligned}\text{constant} &= \boldsymbol{\nabla} \cdot \mathbf{A}(\mathbf{x}, t) \\ &= (2 \pi)^{-3/2} \int i \mathbf{k} \cdot \mathbf{A}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned}
How can this constraint be enforced? The only obvious way is a requirement for $\mathbf{k} \cdot \mathbf{A}(\mathbf{k}, t)$ to be zero for all $(\mathbf{k},t)$, meaning that our to be determined Fourier transform coefficients are required to be perpendicular to the wave number vector parameters at all times.
The remainder of Maxwell’s equations, (3.23) impose the addition constraint on the Fourier transform $\mathbf{A}(\mathbf{k},t)$
\begin{aligned}0 &= (2 \pi)^{-3/2} \int \left( \frac{1}{{c^2}} \partial_{tt} \mathbf{A}(\mathbf{k}, t) - i^2 \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t)\right) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.30)
For zero equality for all $\mathbf{x}$ it appears that we require the Fourier transforms $\mathbf{A}(\mathbf{k})$ to be harmonic in time
\begin{aligned}\partial_{tt} \mathbf{A}(\mathbf{k}, t) = - c^2 \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t).\end{aligned} \hspace{\stretch{1}}(4.31)
This has the familiar exponential solutions
\begin{aligned}\mathbf{A}(\mathbf{k}, t) = \mathbf{A}_{\pm}(\mathbf{k}) e^{ \pm i c {\left\lvert{\mathbf{k}}\right\rvert} t },\end{aligned} \hspace{\stretch{1}}(4.32)
also subject to a requirement that $\mathbf{k} \cdot \mathbf{A}(\mathbf{k}) = 0$. Our field, where the $\mathbf{A}_{\pm}(\mathbf{k})$ are to be determined by initial time conditions, is by (2.7) of the form
\begin{aligned}F(\mathbf{x}, t)= \text{Real} \frac{i}{(\sqrt{2\pi})^3} \int \Bigl( -{\left\lvert{\mathbf{k}}\right\rvert} \mathbf{A}_{+}(\mathbf{k}) + \mathbf{k} \wedge \mathbf{A}_{+}(\mathbf{k}) \Bigr) \exp(i \mathbf{k} \cdot \mathbf{x} + i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}+ \text{Real} \frac{i}{(\sqrt{2\pi})^3} \int \Bigl( {\left\lvert{\mathbf{k}}\right\rvert} \mathbf{A}_{-}(\mathbf{k}) + \mathbf{k} \wedge \mathbf{A}_{-}(\mathbf{k}) \Bigr) \exp(i \mathbf{k} \cdot \mathbf{x} - i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.33)
Since $0 = \mathbf{k} \cdot \mathbf{A}_{\pm}(\mathbf{k})$, we have $\mathbf{k} \wedge \mathbf{A}_{\pm}(\mathbf{k}) = \mathbf{k} \mathbf{A}_{\pm}$. This allows for factoring out of ${\left\lvert{\mathbf{k}}\right\rvert}$. The structure of the solution is not changed by incorporating the $i (2\pi)^{-3/2} {\left\lvert{\mathbf{k}}\right\rvert}$ factors into $\mathbf{A}_{\pm}$, leaving the field having the general form
\begin{aligned}F(\mathbf{x}, t)= \text{Real} \int ( \hat{\mathbf{k}} - 1 ) \mathbf{A}_{+}(\mathbf{k}) \exp(i \mathbf{k} \cdot \mathbf{x} + i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}+ \text{Real} \int ( \hat{\mathbf{k}} + 1 ) \mathbf{A}_{-}(\mathbf{k}) \exp(i \mathbf{k} \cdot \mathbf{x} - i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.34)
The original meaning of $\mathbf{A}_{\pm}$ as Fourier transforms of the vector potential is obscured by the tidy up change to absorb ${\left\lvert{\mathbf{k}}\right\rvert}$, but the geometry of the solution is clearer this way.
It is also particularly straightforward to confirm that $\gamma_0 \nabla F = 0$ separately for either half of (4.34).
## Case III. Non-zero scalar potential. No gauge transformation.
Now lets work from (3.11). In particular, a divergence operation can be factored from (3.11), for
\begin{aligned}0 = \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \phi + \partial_0 \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(4.35)
Right off the top, there is a requirement for
\begin{aligned}\text{constant} = \boldsymbol{\nabla} \phi + \partial_0 \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(4.36)
In terms of the Fourier transforms this is
\begin{aligned}\text{constant} = \frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(i \mathbf{k} \phi(\mathbf{k}, t) + \frac{1}{c} \partial_t \mathbf{A}(\mathbf{k}, t)\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.37)
Are there any ways for this to equal a constant for all $\mathbf{x}$ without requiring that constant to be zero? Assuming no for now, and that this constant must be zero, this implies a coupling between the $\phi$ and $\mathbf{A}$ Fourier transforms of the form
\begin{aligned}\phi(\mathbf{k}, t) = -\frac{1}{{i c \mathbf{k}}} \partial_t \mathbf{A}(\mathbf{k}, t)\end{aligned} \hspace{\stretch{1}}(4.38)
A secondary implication is that $\partial_t \mathbf{A}(\mathbf{k}, t) \propto \mathbf{k}$ or else $\phi(\mathbf{k}, t)$ is not a scalar. We had a transverse solution by requiring via gauge transformation that $\phi = 0$, and here we have instead the vector potential in the propagation direction.
A secondary confirmation that this is a required coupling between the scalar and vector potential can be had by evaluating the divergence equation of (4.35)
\begin{aligned}0 = \frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(- \mathbf{k}^2 \phi(\mathbf{k}, t) + \frac{i\mathbf{k}}{c} \cdot \partial_t \mathbf{A}(\mathbf{k}, t)\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.39)
Rearranging this also produces (4.38). We want to now substitute this relationship into (3.12).
Starting with just the $\partial_0 \phi - \boldsymbol{\nabla} \cdot \mathbf{A}$ part we have
\begin{aligned}\partial_0 \phi + \boldsymbol{\nabla} \cdot \mathbf{A}&=\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(\frac{i}{c^2 \mathbf{k}} \partial_{tt} \mathbf{A}(\mathbf{k}, t) + i \mathbf{k} \cdot \mathbf{A}\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.40)
Taking the gradient of this brings down a factor of $i\mathbf{k}$ for
\begin{aligned}\boldsymbol{\nabla} (\partial_0 \phi + \boldsymbol{\nabla} \cdot \mathbf{A})&=-\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(\frac{1}{c^2} \partial_{tt} \mathbf{A}(\mathbf{k}, t) + \mathbf{k} (\mathbf{k} \cdot \mathbf{A})\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.41)
(3.12) in its entirety is now
\begin{aligned}0 &=\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(- (i\mathbf{k})^2 \mathbf{A}+ \mathbf{k} (\mathbf{k} \cdot \mathbf{A})\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.42)
This isn’t terribly pleasant looking. Perhaps going the other direction. We could write
\begin{aligned}\phi = \frac{i}{c \mathbf{k}} \frac{\partial {\mathbf{A}}}{\partial {t}} = \frac{i}{c} \frac{\partial {\psi}}{\partial {t}},\end{aligned} \hspace{\stretch{1}}(4.43)
so that
\begin{aligned}\mathbf{A}(\mathbf{k}, t) = \mathbf{k} \psi(\mathbf{k}, t).\end{aligned} \hspace{\stretch{1}}(4.44)
\begin{aligned}0 &=\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(\frac{1}{{c^2}} \mathbf{k} \psi_{tt}- \boldsymbol{\nabla}^2 \mathbf{k} \psi + \boldsymbol{\nabla} \frac{i}{c^2} \psi_{tt}+\boldsymbol{\nabla}( \boldsymbol{\nabla} \cdot (\mathbf{k} \psi) )\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \\ \end{aligned}
Note that the gradients here operate on everything to the right, including and especially the exponential. Each application of the gradient brings down an additional $i\mathbf{k}$ factor, and we have
\begin{aligned}\frac{1}{{(\sqrt{2 \pi})^3}} \int \mathbf{k} \Bigl(\frac{1}{{c^2}} \psi_{tt}- i^2 \mathbf{k}^2 \psi + \frac{i^2}{c^2} \psi_{tt}+i^2 \mathbf{k}^2 \psi \Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned}
This is identically zero, so we see that this second equation provides no additional information. That is somewhat surprising since there is not a whole lot of constraints supplied by the first equation. The function $\psi(\mathbf{k}, t)$ can be anything. Understanding of this curiosity comes from computation of the Faraday bivector itself. From (2.7), that is
\begin{aligned}F = \frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(-i \mathbf{k} \frac{i}{c}\psi_t - \frac{1}{c} \mathbf{k} \psi_t + i \mathbf{k} \wedge \mathbf{k} \psi\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.45)
All terms cancel, so we see that a non-zero $\phi$ leads to $F = 0$, as was the case when considering (4.24) (a case that also resulted in $\mathbf{A}(\mathbf{k}) \propto \mathbf{k}$).
Can this Fourier representation lead to a non-transverse solution to Maxwell’s equation? If so, it is not obvious how.
# The energy momentum tensor
The energy momentum tensor is then
\begin{aligned}T(a) &= -\frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left(- \frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)+ i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)- i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)a\left(- \frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)- i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(5.46)
Observing that $\gamma_0$ commutes with spatial bivectors and anticommutes with spatial vectors, and writing $\sigma_\mu = \gamma_\mu \gamma_0$, the tensor splits neatly into scalar and spatial vector components
\begin{aligned}T(\gamma_\mu) \cdot \gamma_0 &= \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left\langle{{\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)+ i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)\sigma_\mu\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)}}\right\rangle e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}' \\ T(\gamma_\mu) \wedge \gamma_0 &= \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint{\left\langle{{\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)+ i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)\sigma_\mu\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)}}\right\rangle}_{1}e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(5.47)
In particular for $\mu = 0$, we have
\begin{aligned}H &\equiv T(\gamma_0) \cdot \gamma_0 = \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left(\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)\right)\cdot\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)\right)- (\mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)) \cdot (\mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t))\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}' \\ \mathbf{P} &\equiv T(\gamma_\mu) \wedge \gamma_0 = \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left(i\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)\right) \cdot\left(\mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)-i\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)\right)\cdot\left(\mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(5.49)
Integrating this over all space and identification of the delta function
\begin{aligned}\delta(\mathbf{k}) \equiv \frac{1}{{(2 \pi)^3}} \int e^{i \mathbf{k} \cdot \mathbf{x}} d^3 \mathbf{x},\end{aligned} \hspace{\stretch{1}}(5.51)
reduces the tensor to a single integral in the continuous angular wave number space of $\mathbf{k}$.
\begin{aligned}\int T(a) d^3 \mathbf{x} &= -\frac{\epsilon_0}{2} \text{Real} \int\left(- \frac{1}{c} {{\dot{\mathbf{A}}}}^{*}+ i \mathbf{k} {{\phi}}^{*}- i \mathbf{k} \wedge {\mathbf{A}}^{*}\right)a\left(- \frac{1}{c} \dot{\mathbf{A}}- i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.52)
Or,
\begin{aligned}\int T(\gamma_\mu) \gamma_0 d^3 \mathbf{x} =\frac{\epsilon_0}{2} \text{Real} \int{\left\langle{{\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}- i \mathbf{k} {{\phi}}^{*}+ i \mathbf{k} \wedge {\mathbf{A}}^{*}\right)\sigma_\mu\left(\frac{1}{c} \dot{\mathbf{A}}+ i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)}}\right\rangle}_{{0,1}}d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.53)
Multiplying out (5.53) yields for $\int H$
\begin{aligned}\int H d^3 \mathbf{x} &=\frac{\epsilon_0}{2} \int d^3 \mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}}\right\rvert}^2 + \mathbf{k}^2 ({\left\lvert{\phi}\right\rvert}^2 + {\left\lvert{\mathbf{A}}\right\rvert}^2 )- {\left\lvert{\mathbf{k} \cdot \mathbf{A}}\right\rvert}^2+ 2 \frac{\mathbf{k}}{c} \cdot \text{Real}( i {{\phi}}^{*} \dot{\mathbf{A}} )\right)\end{aligned} \hspace{\stretch{1}}(5.54)
Recall that the only non-trivial solution we found for the assumed Fourier transform representation of $F$ was for $\phi = 0$, $\mathbf{k} \cdot \mathbf{A}(\mathbf{k}, t) = 0$. Thus we have for the energy density integrated over all space, just
\begin{aligned}\int H d^3 \mathbf{x} &=\frac{\epsilon_0}{2} \int d^3 \mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}}\right\rvert}^2 + \mathbf{k}^2 {\left\lvert{\mathbf{A}}\right\rvert}^2 \right).\end{aligned} \hspace{\stretch{1}}(5.55)
Observe that we have the structure of a Harmonic oscillator for the energy of the radiation system. What is the canonical momentum for this system? Will it correspond to the Poynting vector, integrated over all space?
Let’s reduce the vector component of (5.53), after first imposing the $\phi=0$, and $\mathbf{k} \cdot \mathbf{A} = 0$ conditions used to above for our harmonic oscillator form energy relationship. This is
\begin{aligned}\int \mathbf{P} d^3 \mathbf{x} &=\frac{\epsilon_0}{2 c} \text{Real} \int d^3 \mathbf{k} \left( i {\mathbf{A}}^{*}_t \cdot (\mathbf{k} \wedge \mathbf{A})+ i (\mathbf{k} \wedge {\mathbf{A}}^{*}) \cdot \mathbf{A}_t\right) \\ &=\frac{\epsilon_0}{2 c} \text{Real} \int d^3 \mathbf{k} \left( -i ({\mathbf{A}}^{*}_t \cdot \mathbf{A}) \mathbf{k}+ i \mathbf{k} ({\mathbf{A}}^{*} \cdot \mathbf{A}_t)\right)\end{aligned}
This is just
\begin{aligned}\int \mathbf{P} d^3 \mathbf{x} &=\frac{\epsilon_0}{c} \text{Real} i \int \mathbf{k} ({\mathbf{A}}^{*} \cdot \mathbf{A}_t) d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.56)
Recall that the Fourier transforms for the transverse propagation case had the form $\mathbf{A}(\mathbf{k}, t) = \mathbf{A}_{\pm}(\mathbf{k}) e^{\pm i c {\left\lvert{\mathbf{k}}\right\rvert} t}$, where the minus generated the advanced wave, and the plus the receding wave. With substitution of the vector potential for the advanced wave into the energy and momentum results of (5.55) and (5.56) respectively, we have
\begin{aligned}\int H d^3 \mathbf{x} &= \epsilon_0 \int \mathbf{k}^2 {\left\lvert{\mathbf{A}(\mathbf{k})}\right\rvert}^2 d^3 \mathbf{k} \\ \int \mathbf{P} d^3 \mathbf{x} &= \epsilon_0 \int \hat{\mathbf{k}} \mathbf{k}^2 {\left\lvert{\mathbf{A}(\mathbf{k})}\right\rvert}^2 d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.57)
After a somewhat circuitous route, this has the relativistic symmetry that is expected. In particular the for the complete $\mu=0$ tensor we have after integration over all space
\begin{aligned}\int T(\gamma_0) \gamma_0 d^3 \mathbf{x} = \epsilon_0 \int (1 + \hat{\mathbf{k}}) \mathbf{k}^2 {\left\lvert{\mathbf{A}(\mathbf{k})}\right\rvert}^2 d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.59)
The receding wave solution would give the same result, but directed as $1 - \hat{\mathbf{k}}$ instead.
Observe that we also have the four divergence conservation statement that is expected
\begin{aligned}\frac{\partial {}}{\partial {t}} \int H d^3 \mathbf{x} + \boldsymbol{\nabla} \cdot \int c \mathbf{P} d^3 \mathbf{x} &= 0.\end{aligned} \hspace{\stretch{1}}(5.60)
This follows trivially since both the derivatives are zero. If the integration region was to be more specific instead of a $0 + 0 = 0$ relationship, we’d have the power flux ${\partial {H}}/{\partial {t}}$ equal in magnitude to the momentum change through a bounding surface. For a more general surface the time and spatial dependencies shouldn’t necessarily vanish, but we should still have this radiation energy momentum conservation.
# References
[1] Peeter Joot. Electrodynamic field energy for vacuum. [online]. http://sites.google.com/site/peeterjoot/math2009/fourierMaxVac.pdf.
[2] Peeter Joot. {Energy and momentum for Complex electric and magnetic field phasors.} [online]. http://sites.google.com/site/peeterjoot/math2009/complexFieldEnergy.pdf.
[3] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.
## Energy and momentum for assumed Fourier transform solutions to the homogeneous Maxwell equation.
Posted by peeterjoot on December 22, 2009
# Motivation and notation.
In Electrodynamic field energy for vacuum (reworked) [1], building on Energy and momentum for Complex electric and magnetic field phasors [2] a derivation for the energy and momentum density was derived for an assumed Fourier series solution to the homogeneous Maxwell’s equation. Here we move to the continuous case examining Fourier transform solutions and the associated energy and momentum density.
A complex (phasor) representation is implied, so taking real parts when all is said and done is required of the fields. For the energy momentum tensor the Geometric Algebra form, modified for complex fields, is used
\begin{aligned}T(a) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \hspace{\stretch{1}}(1.1)
The assumed four vector potential will be written
\begin{aligned}A(\mathbf{x}, t) = A^\mu(\mathbf{x}, t) \gamma_\mu = \frac{1}{{(\sqrt{2 \pi})^3}} \int A(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(1.2)
Subject to the requirement that $A$ is a solution of Maxwell’s equation
\begin{aligned}\nabla (\nabla \wedge A) = 0.\end{aligned} \hspace{\stretch{1}}(1.3)
To avoid latex hell, no special notation will be used for the Fourier coefficients,
\begin{aligned}A(\mathbf{k}, t) = \frac{1}{{(\sqrt{2 \pi})^3}} \int A(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{x}.\end{aligned} \hspace{\stretch{1}}(1.4)
When convenient and unambiguous, this $(\mathbf{k},t)$ dependence will be implied.
Having picked a time and space representation for the field, it will be natural to express both the four potential and the gradient as scalar plus spatial vector, instead of using the Dirac basis. For the gradient this is
\begin{aligned}\nabla &= \gamma^\mu \partial_\mu = (\partial_0 - \boldsymbol{\nabla}) \gamma_0 = \gamma_0 (\partial_0 + \boldsymbol{\nabla}),\end{aligned} \hspace{\stretch{1}}(1.5)
and for the four potential (or the Fourier transform functions), this is
\begin{aligned}A &= \gamma_\mu A^\mu = (\phi + \mathbf{A}) \gamma_0 = \gamma_0 (\phi - \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(1.6)
# Setup
The field bivector $F = \nabla \wedge A$ is required for the energy momentum tensor. This is
\begin{aligned}\nabla \wedge A&= \frac{1}{{2}}\left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \\ &= \frac{1}{{2}}\left( (\stackrel{ \rightarrow }{\partial}_0 - \stackrel{ \rightarrow }{\boldsymbol{\nabla}}) \gamma_0 \gamma_0 (\phi - \mathbf{A})- (\phi + \mathbf{A}) \gamma_0 \gamma_0 (\stackrel{ \leftarrow }{\partial}_0 + \stackrel{ \leftarrow }{\boldsymbol{\nabla}})\right) \\ &= -\boldsymbol{\nabla} \phi -\partial_0 \mathbf{A} + \frac{1}{{2}}(\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{A} - \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}}) \end{aligned}
This last term is a spatial curl and the field is then
\begin{aligned}F = -\boldsymbol{\nabla} \phi -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} \end{aligned} \hspace{\stretch{1}}(2.7)
Applied to the Fourier representation this is
\begin{aligned}F = \frac{1}{{(\sqrt{2 \pi})^3}} \int \left( - \frac{1}{{c}} \dot{\mathbf{A}}- i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(2.8)
The energy momentum tensor is then
\begin{aligned}T(a) &= -\frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint \left( - \frac{1}{{c}} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)+ i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)- i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)a\left( - \frac{1}{{c}} \dot{\mathbf{A}}(\mathbf{k}, t)- i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(2.9)
# The tensor integrated over all space. Energy and momentum?
Integrating this over all space and identification of the delta function
\begin{aligned}\delta(\mathbf{k}) \equiv \frac{1}{{(2 \pi)^3}} \int e^{i \mathbf{k} \cdot \mathbf{x}} d^3 \mathbf{x},\end{aligned} \hspace{\stretch{1}}(3.10)
reduces the tensor to a single integral in the continuous angular wave number space of $\mathbf{k}$.
\begin{aligned}\int T(a) d^3 \mathbf{x} &= -\frac{\epsilon_0}{2} \text{Real} \int \left( - \frac{1}{{c}} {{\dot{\mathbf{A}}}}^{*}+ i \mathbf{k} {{\phi}}^{*}- i \mathbf{k} \wedge {\mathbf{A}}^{*}\right)a\left( - \frac{1}{{c}} \dot{\mathbf{A}}- i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(3.11)
Observing that $\gamma_0$ commutes with spatial bivectors and anticommutes with spatial vectors, and writing $\sigma_\mu = \gamma_\mu \gamma_0$, one has
\begin{aligned}\int T(\gamma_\mu) \gamma_0 d^3 \mathbf{x} = \frac{\epsilon_0}{2} \text{Real} \int {\left\langle{{\left( \frac{1}{{c}} {{\dot{\mathbf{A}}}}^{*}- i \mathbf{k} {{\phi}}^{*}+ i \mathbf{k} \wedge {\mathbf{A}}^{*}\right)\sigma_\mu\left( \frac{1}{{c}} \dot{\mathbf{A}}+ i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)}}\right\rangle}_{{0,1}}d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(3.12)
The scalar and spatial vector grade selection operator has been added for convenience and does not change the result since those are necessarily the only grades anyhow. The post multiplication by the observer frame time basis vector $\gamma_0$ serves to separate the energy and momentum like components of the tensor nicely into scalar and vector aspects. In particular for $T(\gamma^0)$, one could write
\begin{aligned}\int T(\gamma^0) d^3 \mathbf{x} = (H + \mathbf{P}) \gamma_0,\end{aligned} \hspace{\stretch{1}}(3.13)
If these are correctly identified with energy and momentum then it also ought to be true that we have the conservation relationship
\begin{aligned}\frac{\partial {H}}{\partial {t}} + \boldsymbol{\nabla} \cdot (c \mathbf{P}) = 0.\end{aligned} \hspace{\stretch{1}}(3.14)
However, multiplying out (3.12) yields for $H$
\begin{aligned}H &= \frac{\epsilon_0}{2} \int d^3 \mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}}\right\rvert}^2 + \mathbf{k}^2 ({\left\lvert{\phi}\right\rvert}^2 + {\left\lvert{\mathbf{A}}\right\rvert}^2 )- {\left\lvert{\mathbf{k} \cdot \mathbf{A}}\right\rvert}^2 + 2 \frac{\mathbf{k}}{c} \cdot \text{Real}( i {{\phi}}^{*} \dot{\mathbf{A}} )\right)\end{aligned} \hspace{\stretch{1}}(3.15)
The vector component takes a bit more work to reduce
\begin{aligned}\mathbf{P} &= \frac{\epsilon_0}{2} \int d^3 \mathbf{k} \text{Real} \left(\frac{i}{c} ({{\dot{\mathbf{A}}}}^{*} \cdot (\mathbf{k} \wedge \mathbf{A})+ {{\phi}}^{*} \mathbf{k} \cdot (\mathbf{k} \wedge \mathbf{A})+ \frac{i}{c} (\mathbf{k} \wedge {\mathbf{A}}^{*}) \cdot \dot{\mathbf{A}}- \phi (\mathbf{k} \wedge {\mathbf{A}}^{*}) \cdot \mathbf{k}\right) \\ &=\frac{\epsilon_0}{2} \int d^3 \mathbf{k} \text{Real} \left(\frac{i}{c} \left( ({{\dot{\mathbf{A}}}}^{*} \cdot \mathbf{k}) \mathbf{A} -({{\dot{\mathbf{A}}}}^{*} \cdot \mathbf{A}) \mathbf{k} \right)+ {{\phi}}^{*} \left( \mathbf{k}^2 \mathbf{A} - (\mathbf{k} \cdot \mathbf{A}) \mathbf{k} \right)+ \frac{i}{c} \left( ({\mathbf{A}}^{*} \cdot \dot{\mathbf{A}}) \mathbf{k} - (\mathbf{k} \cdot \dot{\mathbf{A}}) {\mathbf{A}}^{*} \right)+ \phi \left( \mathbf{k}^2 {\mathbf{A}}^{*} -({\mathbf{A}}^{*} \cdot \mathbf{k}) \mathbf{k} \right) \right).\end{aligned}
Canceling and regrouping leaves
\begin{aligned}\mathbf{P}&=\epsilon_0 \int d^3 \mathbf{k} \text{Real} \left(\mathbf{A} \left( \mathbf{k}^2 {{\phi}}^{*} + \mathbf{k} \cdot {{\dot{\mathbf{A}}}}^{*} \right)+ \mathbf{k} \left( -{{\phi}}^{*} (\mathbf{k} \cdot \mathbf{A}) + \frac{i}{c} ({\mathbf{A}}^{*} \cdot \dot{\mathbf{A}})\right)\right).\end{aligned} \hspace{\stretch{1}}(3.16)
This has no explicit $\mathbf{x}$ dependence, so the conservation relation (3.14) is violated unless ${\partial {H}}/{\partial {t}} = 0$. There is no reason to assume that will be the case. In the discrete Fourier series treatment, a gauge transformation allowed for elimination of $\phi$, and this implied $\mathbf{k} \cdot \mathbf{A}_\mathbf{k} = 0$ or $\mathbf{A}_\mathbf{k}$ constant. We will probably have a similar result here, eliminating most of the terms in (3.15) and (3.16). Except for the constant $\mathbf{A}_\mathbf{k}$ solution of the field equations there is no obvious way that such a simplified energy expression will have zero derivative.
A more reasonable conclusion is that this approach is flawed. We ought to be looking at the divergence relation as a starting point, and instead of integrating over all space, instead employing Gauss’s theorem to convert the divergence integral into a surface integral. Without math, the conservation relationship probably ought to be expressed as energy change in a volume is matched by the momentum change through the surface. However, without an integral over all space, we do not get the nice delta function cancellation observed above. How to proceed is not immediately clear. Stepping back to review applications of Gauss’s theorem is probably a good first step.
# References
[1] Peeter Joot. Electrodynamic field energy for vacuum. [online]. http://sites.google.com/site/peeterjoot/math2009/fourierMaxVac.pdf.
[2] Peeter Joot. {Energy and momentum for Complex electric and magnetic field phasors.} [online]. http://sites.google.com/site/peeterjoot/math2009/complexFieldEnergy.pdf.
## Electrodynamic field energy for vacuum (reworked)
Posted by peeterjoot on December 21, 2009
# Previous version.
Reducing the products in the Dirac basis makes life more complicated then it needs to be (became obvious when attempting to derive an expression for the Poynting integral).
# Motivation.
From Energy and momentum for Complex electric and magnetic field phasors [PDF] how to formulate the energy momentum tensor for complex vector fields (ie. phasors) in the Geometric Algebra formalism is now understood. To recap, for the field $F = \mathbf{E} + I c \mathbf{B}$, where $\mathbf{E}$ and $\mathbf{B}$ may be complex vectors we have for Maxwell’s equation
\begin{aligned}\nabla F = J/\epsilon_0 c.\end{aligned} \quad\quad\quad(1)
This is a doubly complex representation, with the four vector pseudoscalar $I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$ acting as a non-commutatitive imaginary, as well as real and imaginary parts for the electric and magnetic field vectors. We take the real part (not the scalar part) of any bivector solution $F$ of Maxwell’s equation as the actual solution, but allow ourself the freedom to work with the complex phasor representation when convenient. In these phasor vectors, the imaginary $i$, as in $\mathbf{E} = \text{Real}(\mathbf{E}) + i \text{Imag}(\mathbf{E})$, is a commuting imaginary, commuting with all the multivector elements in the algebra.
The real valued, four vector, energy momentum tensor $T(a)$ was found to be
\begin{aligned}T(a) = \frac{\epsilon_0}{4} \Bigl( {{F}}^{*} a \tilde{F} + \tilde{F} a {{F}}^{*} \Bigr) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \quad\quad\quad(2)
To supply some context that gives meaning to this tensor the associated conservation relationship was found to be
\begin{aligned}\nabla \cdot T(a) &= a \cdot \frac{1}{{ c }} \text{Real} \left( J \cdot {{F}}^{*} \right).\end{aligned} \quad\quad\quad(3)
and in particular for $a = \gamma^0$, this four vector divergence takes the form
\begin{aligned}\frac{\partial {}}{\partial {t}}\frac{\epsilon_0}{2}(\mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*})+ \boldsymbol{\nabla} \cdot \frac{1}{{\mu_0}} \text{Real} (\mathbf{E} \times {\mathbf{B}}^{*} )+ \text{Real}( \mathbf{J} \cdot {\mathbf{E}}^{*} ) = 0,\end{aligned} \quad\quad\quad(4)
relating the energy term $T^{00} = T(\gamma^0) \cdot \gamma^0$ and the Poynting spatial vector $T(\gamma^0) \wedge \gamma^0$ with the current density and electric field product that constitutes the energy portion of the Lorentz force density.
Let’s apply this to calculating the energy associated with the field that is periodic within a rectangular prism as done by Bohm in [2]. We do not necessarily need the Geometric Algebra formalism for this calculation, but this will be a fun way to attempt it.
# Setup
Let’s assume a Fourier representation for the four vector potential $A$ for the field $F = \nabla \wedge A$. That is
\begin{aligned}A = \sum_{\mathbf{k}} A_\mathbf{k}(t) e^{i \mathbf{k} \cdot \mathbf{x}},\end{aligned} \quad\quad\quad(5)
where summation is over all angular wave number triplets $\mathbf{k} = 2 \pi (k_1/\lambda_1, k_2/\lambda_2, k_3/\lambda_3)$. The Fourier coefficients $A_\mathbf{k} = {A_\mathbf{k}}^\mu \gamma_\mu$ are allowed to be complex valued, as is the resulting four vector $A$, and the associated bivector field $F$.
Fourier inversion, with $V = \lambda_1 \lambda_2 \lambda_3$, follows from
\begin{aligned}\delta_{\mathbf{k}', \mathbf{k}} =\frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} e^{ i \mathbf{k}' \cdot \mathbf{x}} e^{-i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(6)
but only this orthogonality relationship and not the Fourier coefficients themselves
\begin{aligned}A_\mathbf{k} = \frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} A(\mathbf{x}, t) e^{- i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(7)
will be of interest here. Evaluating the curl for this potential yields
\begin{aligned}F = \nabla \wedge A= \sum_{\mathbf{k}} \left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^m \wedge A_\mathbf{k} \frac{2 \pi i k_m}{\lambda_m} \right) e^{i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(8)
Since the four vector potential has been expressed using an explicit split into time and space components it will be natural to re express the bivector field in terms of scalar and (spatial) vector potentials, with the Fourier coefficients. Writing $\sigma_m = \gamma_m \gamma_0$ for the spatial basis vectors, ${A_\mathbf{k}}^0 = \phi_\mathbf{k}$, and $\mathbf{A} = A^k \sigma_k$, this is
\begin{aligned}A_\mathbf{k} = (\phi_\mathbf{k} + \mathbf{A}_\mathbf{k}) \gamma_0.\end{aligned} \quad\quad\quad(9)
The Faraday bivector field $F$ is then
\begin{aligned}F = \sum_\mathbf{k} \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) e^{i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(10)
This is now enough to express the energy momentum tensor $T(\gamma^\mu)$
\begin{aligned}T(\gamma^\mu) &= -\frac{\epsilon_0}{2} \sum_{\mathbf{k},\mathbf{k}'}\text{Real} \left(\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}'})}}^{*} + i \mathbf{k}' {{\phi_{\mathbf{k}'}}}^{*} - i \mathbf{k}' \wedge {{\mathbf{A}_{\mathbf{k}'}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x}}\right).\end{aligned} \quad\quad\quad(11)
It will be more convenient to work with a scalar plus bivector (spatial vector) form of this tensor, and right multiplication by $\gamma_0$ produces such a split
\begin{aligned}T(\gamma^\mu) \gamma_0 = \left\langle{{T(\gamma^\mu) \gamma_0}}\right\rangle + \sigma_a \left\langle{{ \sigma_a T(\gamma^\mu) \gamma_0 }}\right\rangle\end{aligned} \quad\quad\quad(12)
The primary object of this treatment will be consideration of the $\mu = 0$ components of the tensor, which provide a split into energy density $T(\gamma^0) \cdot \gamma_0$, and Poynting vector (momentum density) $T(\gamma^0) \wedge \gamma_0$.
Our first step is to integrate (12) over the volume $V$. This integration and the orthogonality relationship (6), removes the exponentials, leaving
\begin{aligned}\int T(\gamma^\mu) \cdot \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) \gamma_0 }}\right\rangle \\ \int T(\gamma^\mu) \wedge \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) \gamma_0}}\right\rangle \end{aligned} \quad\quad\quad(13)
Because $\gamma_0$ commutes with the spatial bivectors, and anticommutes with the spatial vectors, the remainder of the Dirac basis vectors in these expressions can be eliminated
\begin{aligned}\int T(\gamma^0) \cdot \gamma_0&= -\frac{\epsilon_0 V }{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(15)
\begin{aligned}\int T(\gamma^0) \wedge \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(16)
\begin{aligned}\int T(\gamma^m) \cdot \gamma_0&= \frac{\epsilon_0 V }{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \sigma_m\left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(17)
\begin{aligned}\int T(\gamma^m) \wedge \gamma_0&= \frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \sigma_m\left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle.\end{aligned} \quad\quad\quad(18)
# Expanding the energy momentum tensor components.
## Energy
In (15) only the bivector-bivector and vector-vector products produce any scalar grades. Except for the bivector product this can be done by inspection. For that part we utilize the identity
\begin{aligned}\left\langle{{ (\mathbf{k} \wedge \mathbf{a}) (\mathbf{k} \wedge \mathbf{b}) }}\right\rangle= (\mathbf{a} \cdot \mathbf{k}) (\mathbf{b} \cdot \mathbf{k}) - \mathbf{k}^2 (\mathbf{a} \cdot \mathbf{b}).\end{aligned} \quad\quad\quad(19)
This leaves for the energy $H = \int T(\gamma^0) \cdot \gamma_0$ in the volume
\begin{aligned}H = \frac{\epsilon_0 V}{2} \sum_\mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2 +\mathbf{k}^2 \left( {\left\lvert{\phi_\mathbf{k}}\right\rvert}^2 + {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right) - {\left\lvert{\mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2+ \frac{2}{c} \text{Real} \left( i {{\phi_\mathbf{k}}}^{*} \cdot \dot{\mathbf{A}}_\mathbf{k} \right)\right)\end{aligned} \quad\quad\quad(20)
We are left with a completely real expression, and one without any explicit Geometric Algebra. This does not look like the Harmonic oscillator Hamiltonian that was expected. A gauge transformation to eliminate $\phi_\mathbf{k}$ and an observation about when $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ equals zero will give us that, but first lets get the mechanical jobs done, and reduce the products for the field momentum.
## Momentum
Now move on to (16). For the factors other than $\sigma_a$ only the vector-bivector products can contribute to the scalar product. We have two such products, one of the form
\begin{aligned}\sigma_a \left\langle{{ \sigma_a \mathbf{a} (\mathbf{k} \wedge \mathbf{c}) }}\right\rangle&=\sigma_a (\mathbf{c} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{k}) - \sigma_a (\mathbf{k} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{c}) \\ &=\mathbf{c} (\mathbf{a} \cdot \mathbf{k}) - \mathbf{k} (\mathbf{a} \cdot \mathbf{c}),\end{aligned}
and the other
\begin{aligned}\sigma_a \left\langle{{ \sigma_a (\mathbf{k} \wedge \mathbf{c}) \mathbf{a} }}\right\rangle&=\sigma_a (\mathbf{k} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{c}) - \sigma_a (\mathbf{c} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{k}) \\ &=\mathbf{k} (\mathbf{a} \cdot \mathbf{c}) - \mathbf{c} (\mathbf{a} \cdot \mathbf{k}).\end{aligned}
The momentum $\mathbf{P} = \int T(\gamma^0) \wedge \gamma_0$ in this volume follows by computation of
\begin{aligned}&\sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \\ &= i \mathbf{A}_\mathbf{k} \left( \left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} \right) \cdot \mathbf{k} \right) - i \mathbf{k} \left( \left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} \right) \cdot \mathbf{A}_\mathbf{k} \right) \\ &- i \mathbf{k} \left( \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} \right) \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right) + i {{\mathbf{A}_{\mathbf{k}}}}^{*} \left( \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} \right) \cdot \mathbf{k} \right)\end{aligned}
All the products are paired in nice conjugates, taking real parts, and premultiplication with $-\epsilon_0 V/2$ gives the desired result. Observe that two of these terms cancel, and another two have no real part. Those last are
\begin{aligned}-\frac{\epsilon_0 V \mathbf{k}}{2 c} \text{Real} \left( i {{(\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{A}_\mathbf{k}+\dot{\mathbf{A}}_\mathbf{k} \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right)&=-\frac{\epsilon_0 V \mathbf{k}}{2 c} \text{Real} \left( i \frac{d}{dt} \mathbf{A}_\mathbf{k} \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right)\end{aligned}
Taking the real part of this pure imaginary $i {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2$ is zero, leaving just
\begin{aligned}\mathbf{P} &= \epsilon_0 V \sum_{\mathbf{k}}\text{Real} \left(i \mathbf{A}_\mathbf{k} \left( \frac{1}{{c}} {{\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{k} \right)+ \mathbf{k}^2 \phi_\mathbf{k} {{ \mathbf{A}_\mathbf{k} }}^{*}- \mathbf{k} {{\phi_\mathbf{k}}}^{*} (\mathbf{k} \cdot \mathbf{A}_\mathbf{k})\right)\end{aligned} \quad\quad\quad(21)
I am not sure why exactly, but I actually expected a term with ${\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2$, quadratic in the vector potential. Is there a mistake above?
## Gauge transformation to simplify the Hamiltonian.
In (20) something that looked like the Harmonic oscillator was expected. On the surface this does not appear to be such a beast. Exploitation of gauge freedom is required to make the simplification that puts things into the Harmonic oscillator form.
If we are to change our four vector potential $A \rightarrow A + \nabla \psi$, then Maxwell’s equation takes the form
\begin{aligned}J/\epsilon_0 c = \nabla (\nabla \wedge (A + \nabla \psi) = \nabla (\nabla \wedge A) + \nabla (\underbrace{\nabla \wedge \nabla \psi}_{=0}),\end{aligned} \quad\quad\quad(22)
which is unchanged by the addition of the gradient to any original potential solution to the equation. In coordinates this is a transformation of the form
\begin{aligned}A^\mu \rightarrow A^\mu + \partial_\mu \psi,\end{aligned} \quad\quad\quad(23)
and we can use this to force any one of the potential coordinates to zero. For this problem, it appears that it is desirable to seek a $\psi$ such that $A^0 + \partial_0 \psi = 0$. That is
\begin{aligned}\sum_\mathbf{k} \phi_\mathbf{k}(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \frac{1}{{c}} \partial_t \psi = 0.\end{aligned} \quad\quad\quad(24)
Or,
\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x},0) -\frac{1}{{c}} \sum_\mathbf{k} e^{i \mathbf{k} \cdot \mathbf{x}} \int_{\tau=0}^t \phi_\mathbf{k}(\tau).\end{aligned} \quad\quad\quad(25)
With such a transformation, the $\phi_\mathbf{k}$ and $\dot{\mathbf{A}}_\mathbf{k}$ cross term in the Hamiltonian (20) vanishes, as does the $\phi_\mathbf{k}$ term in the four vector square of the last term, leaving just
\begin{aligned}H = \frac{\epsilon_0}{c^2} V \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} \Bigl((c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 + {\left\lvert{ ( c \mathbf{k}) \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2+ {\left\lvert{ c \mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(26)
Additionally, wedging (5) with $\gamma_0$ now does not loose any information so our potential Fourier series is reduced to just
\begin{aligned}\mathbf{A} &= \sum_{\mathbf{k}} \mathbf{A}_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \\ \mathbf{A}_\mathbf{k} &= \frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} \mathbf{A}(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3.\end{aligned} \quad\quad\quad(27)
The desired harmonic oscillator form would be had in (26) if it were not for the $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ term. Does that vanish? Returning to Maxwell’s equation should answer that question, but first it has to be expressed in terms of the vector potential. While $\mathbf{A} = A \wedge \gamma_0$, the lack of an $A^0$ component means that this can be inverted as
\begin{aligned}A = \mathbf{A} \gamma_0 = -\gamma_0 \mathbf{A}.\end{aligned} \quad\quad\quad(29)
The gradient can also be factored scalar and spatial vector components
\begin{aligned}\nabla = \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) = ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0.\end{aligned} \quad\quad\quad(30)
So, with this $A^0 = 0$ gauge choice the bivector field $F$ is
\begin{aligned}F = \nabla \wedge A = \frac{1}{{2}} \left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \end{aligned} \quad\quad\quad(31)
From the left the gradient action on $A$ is
\begin{aligned}\stackrel{ \rightarrow }{\nabla} A &= ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0 (-\gamma_0 \mathbf{A}) \\ &= ( -\partial_0 + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} ) \mathbf{A},\end{aligned}
and from the right
\begin{aligned}A \stackrel{ \leftarrow }{\nabla}&= \mathbf{A} \gamma_0 \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \mathbf{A} ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \partial_0 \mathbf{A} + \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \end{aligned}
Taking the difference we have
\begin{aligned}F &= \frac{1}{{2}} \Bigl( -\partial_0 \mathbf{A} + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{A} - \partial_0 \mathbf{A} - \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \Bigr).\end{aligned}
Which is just
\begin{aligned}F = -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A}.\end{aligned} \quad\quad\quad(32)
For this vacuum case, premultiplication of Maxwell’s equation by $\gamma_0$ gives
\begin{aligned}0 &= \gamma_0 \nabla ( -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &= (\partial_0 + \boldsymbol{\nabla})( -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} - \partial_0 \boldsymbol{\nabla} \cdot \mathbf{A} - \partial_0 \boldsymbol{\nabla} \wedge \mathbf{A} + \partial_0 ( \boldsymbol{\nabla} \wedge \mathbf{A} ) + \underbrace{\boldsymbol{\nabla} \cdot ( \boldsymbol{\nabla} \wedge \mathbf{A} ) }_{\boldsymbol{\nabla}^2 \mathbf{A} - \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A})}+ \underbrace{\boldsymbol{\nabla} \wedge ( \boldsymbol{\nabla} \wedge \mathbf{A} )}_{=0} \\ \end{aligned}
The spatial bivector and trivector grades are all zero. Equating the remaining scalar and vector components to zero separately yields a pair of equations in $\mathbf{A}$
\begin{aligned}0 &= \partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) \\ 0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) \end{aligned} \quad\quad\quad(33)
If the divergence of the vector potential is constant we have just a wave equation. Let’s see what that divergence is with the assumed Fourier representation
\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{A} &=\sum_{\mathbf{k} \ne (0,0,0)} {\mathbf{A}_\mathbf{k}}^m 2 \pi i \frac{k_m}{\lambda_m} e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=i \sum_{\mathbf{k} \ne (0,0,0)} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=i \sum_\mathbf{k} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} \end{aligned}
Since $\mathbf{A}_\mathbf{k} = \mathbf{A}_\mathbf{k}(t)$, there are two ways for $\partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) = 0$. For each $\mathbf{k}$ there must be a requirement for either $\mathbf{A}_\mathbf{k} \cdot \mathbf{k} = 0$ or $\mathbf{A}_\mathbf{k} = \text{constant}$. The constant $\mathbf{A}_\mathbf{k}$ solution to the first equation appears to represent a standing spatial wave with no time dependence. Is that of any interest?
The more interesting seeming case is where we have some non-static time varying state. In this case, if $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$, the second of these Maxwell’s equations is just the vector potential wave equation, since the divergence is zero. That is
\begin{aligned}0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} \end{aligned} \quad\quad\quad(35)
Solving this isn’t really what is of interest, since the objective was just to determine if the divergence could be assumed to be zero. This shows then, that if the transverse solution to Maxwell’s equation is picked, the Hamiltonian for this field, with this gauge choice, becomes
\begin{aligned}H = \frac{\epsilon_0}{c^2} V \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} (c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right).\end{aligned} \quad\quad\quad(36)
How does the gauge choice alter the Poynting vector? From (21), all the $\phi_\mathbf{k}$ dependence in that integrated momentum density is lost
\begin{aligned}\mathbf{P} &= \epsilon_0 V \sum_{\mathbf{k}}\text{Real} \left(i \mathbf{A}_\mathbf{k} \left( \frac{1}{{c}} {{\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{k} \right)\right).\end{aligned} \quad\quad\quad(37)
The $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$ solutions to Maxwell’s equation are seen to result in zero momentum for this infinite periodic field. My expectation was something of the form $c \mathbf{P} = H \hat{\mathbf{k}}$, so intuition is either failing me, or my math is failing me, or this contrived periodic field solution leads to trouble.
# Conclusions and followup.
The objective was met, a reproduction of Bohm’s Harmonic oscillator result using a complex exponential Fourier series instead of separate sine and cosines.
The reason for Bohm’s choice to fix zero divergence as the gauge choice upfront is now clear. That automatically cuts complexity from the results. Figuring out how to work this problem with complex valued potentials and also using the Geometric Algebra formulation probably also made the work a bit more difficult since blundering through both simultaneously was required instead of just one at a time.
This was an interesting exercise though, since doing it this way I am able to understand all the intermediate steps. Bohm employed some subtler argumentation to eliminate the scalar potential $\phi$ upfront, and I have to admit I did not follow his logic, whereas blindly following where the math leads me all makes sense.
As a bit of followup, I’d like to consider the constant $\mathbf{A}_\mathbf{k}$ case in more detail, and any implications of the freedom to pick $\mathbf{A}_0$.
The general calculation of $T^{\mu\nu}$ for the assumed Fourier solution should be possible too, but was not attempted. Doing that general calculation with a four dimensional Fourier series is likely tidier than working with scalar and spatial variables as done here.
Now that the math is out of the way (except possibly for the momentum which doesn’t seem right), some discussion of implications and applications is also in order. My preference is to let the math sink-in a bit first and mull over the momentum issues at leisure.
# References
[2] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.
## Electrodynamic field energy for vacuum.
Posted by peeterjoot on December 19, 2009
# Motivation.
We now know how to formulate the energy momentum tensor for complex vector fields (ie. phasors) in the Geometric Algebra formalism. To recap, for the field $F = \mathbf{E} + I c \mathbf{B}$, where $\mathbf{E}$ and $\mathbf{B}$ may be complex vectors we have for Maxwell’s equation
\begin{aligned}\nabla F = J/\epsilon_0 c.\end{aligned} \quad\quad\quad(1)
This is a doubly complex representation, with the four vector pseudoscalar $I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$ acting as a non-commutatitive imaginary, as well as real and imaginary parts for the electric and magnetic field vectors. We take the real part (not the scalar part) of any bivector solution $F$ of Maxwell’s equation as the actual solution, but allow ourself the freedom to work with the complex phasor representation when convenient. In these phasor vectors, the imaginary $i$, as in $\mathbf{E} = \text{Real}(\mathbf{E}) + i \text{Imag}(\mathbf{E})$, is a commuting imaginary, commuting with all the multivector elements in the algebra.
The real valued, four vector, energy momentum tensor $T(a)$ was found to be
\begin{aligned}T(a) = \frac{\epsilon_0}{4} \Bigl( {{F}}^{*} a \tilde{F} + \tilde{F} a {{F}}^{*} \Bigr) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \quad\quad\quad(2)
To supply some context that gives meaning to this tensor the associated conservation relationship was found to be
\begin{aligned}\nabla \cdot T(a) &= a \cdot \frac{1}{{ c }} \text{Real} \left( J \cdot {{F}}^{*} \right).\end{aligned} \quad\quad\quad(3)
and in particular for $a = \gamma^0$, this four vector divergence takes the form
\begin{aligned}\frac{\partial {}}{\partial {t}}\frac{\epsilon_0}{2}(\mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*})+ \boldsymbol{\nabla} \cdot \frac{1}{{\mu_0}} \text{Real} (\mathbf{E} \times {\mathbf{B}}^{*} )+ \text{Real}( \mathbf{J} \cdot {\mathbf{E}}^{*} ) = 0,\end{aligned} \quad\quad\quad(4)
relating the energy term $T^{00} = T(\gamma^0) \cdot \gamma^0$ and the Poynting spatial vector $T(\gamma^0) \wedge \gamma^0$ with the current density and electric field product that constitutes the energy portion of the Lorentz force density.
Let’s apply this to calculating the energy associated with the field that is periodic within a rectangular prism as done by Bohm in [1]. We do not necessarily need the Geometric Algebra formalism for this calculation, but this will be a fun way to attempt it.
# Setup
Let’s assume a Fourier representation for the four vector potential $A$ for the field $F = \nabla \wedge A$. That is
\begin{aligned}A = \sum_{\mathbf{k}} A_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}},\end{aligned} \quad\quad\quad(5)
where summation is over all wave number triplets $\mathbf{k} = (p/\lambda_1,q/\lambda_2,r/\lambda_3)$. The Fourier coefficients $A_\mathbf{k} = {A_\mathbf{k}}^\mu \gamma_\mu$ are allowed to be complex valued, as is the resulting four vector $A$, and the associated bivector field $F$.
Fourier inversion follows from
\begin{aligned}\delta_{\mathbf{k}', \mathbf{k}} =\frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} e^{2 \pi i \mathbf{k}' \cdot \mathbf{x}} e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(6)
but only this orthogonality relationship and not the Fourier coefficients themselves
\begin{aligned}A_\mathbf{k} = \frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} A(\mathbf{x}, t) e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(7)
will be of interest here. Evaluating the curl for this potential yields
\begin{aligned}F = \nabla \wedge A= \sum_{\mathbf{k}} \left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \sum_{m=1}^3 \gamma^m \wedge A_\mathbf{k} \frac{2 \pi i k_m}{\lambda_m} \right) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(8)
We can now form the energy density
\begin{aligned}U = T(\gamma^0) \cdot \gamma^0=-\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} \gamma^0 F \gamma^0 \Bigr).\end{aligned} \quad\quad\quad(9)
With implied summation over all repeated integer indexes (even without matching uppers and lowers), this is
\begin{aligned}U =-\frac{\epsilon_0}{2} \sum_{\mathbf{k}', \mathbf{k}} \text{Real} \left\langle{{\left( \frac{1}{{c}} \gamma^0 \wedge {{\dot{A}_{\mathbf{k}'}}}^{*} - \gamma^m \wedge {{A_{\mathbf{k}'}}}^{*} \frac{2 \pi i k_m'}{\lambda_m} \right) e^{-2 \pi i \mathbf{k}' \cdot \mathbf{x}}\gamma^0\left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^n \wedge A_\mathbf{k} \frac{2 \pi i k_n}{\lambda_n} \right) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}}\gamma^0}}\right\rangle.\end{aligned} \quad\quad\quad(10)
The grade selection used here doesn’t change the result since we already have a scalar, but will just make it convenient to filter out any higher order products that will cancel anyways. Integrating over the volume element and taking advantage of the orthogonality relationship (6), the exponentials are removed, leaving the energy contained in the volume
\begin{aligned}H = -\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2}\sum_{\mathbf{k}} \text{Real} \left\langle{{\left( \frac{1}{{c}} \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} - \gamma^m \wedge {{A_{\mathbf{k}}}}^{*} \frac{2 \pi i k_m}{\lambda_m} \right) \gamma^0\left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^n \wedge A_\mathbf{k} \frac{2 \pi i k_n}{\lambda_n} \right) \gamma^0}}\right\rangle.\end{aligned} \quad\quad\quad(11)
# First reduction of the Hamiltonian.
Let’s take the products involved in sequence one at a time, and evaluate, later adding and taking real parts if required all of
\begin{aligned}\frac{1}{{c^2}}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) \gamma^0 (\gamma^0 \wedge \dot{A}_\mathbf{k}) \gamma^0 }}\right\rangle &=-\frac{1}{{c^2}}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) (\gamma^0 \wedge \dot{A}_\mathbf{k}) }}\right\rangle \end{aligned} \quad\quad\quad(12)
\begin{aligned}- \frac{2 \pi i k_m}{c \lambda_m} \left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) \gamma^0 ( \gamma^0 \wedge \dot{A}_\mathbf{k} ) \gamma^0}}\right\rangle &=\frac{2 \pi i k_m}{c \lambda_m} \left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) ( \gamma^0 \wedge \dot{A}_\mathbf{k} ) }}\right\rangle \end{aligned} \quad\quad\quad(13)
\begin{aligned}\frac{2 \pi i k_n}{c \lambda_n} \left\langle{{ ( \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) \gamma^0 ( \gamma^n \wedge A_\mathbf{k} ) \gamma^0}}\right\rangle &=-\frac{2 \pi i k_n}{c \lambda_n} \left\langle{{ ( \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) ( \gamma^n \wedge A_\mathbf{k} ) }}\right\rangle \end{aligned} \quad\quad\quad(14)
\begin{aligned}-\frac{4 \pi^2 k_m k_n}{\lambda_m \lambda_n}\left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) \gamma^0(\gamma^n \wedge A_\mathbf{k} ) \gamma^0}}\right\rangle. &\end{aligned} \quad\quad\quad(15)
The expectation is to obtain a Hamiltonian for the field that has the structure of harmonic oscillators, where the middle two products would have to be zero or sum to zero or have real parts that sum to zero. The first is expected to contain only products of ${\left\lvert{{\dot{A}_\mathbf{k}}^m}\right\rvert}^2$, and the last only products of ${\left\lvert{{A_\mathbf{k}}^m}\right\rvert}^2$.
While initially guessing that (13) and (14) may cancel, this isn’t so obviously the case. The use of cyclic permutation of multivectors within the scalar grade selection operator $\left\langle{{A B}}\right\rangle = \left\langle{{B A}}\right\rangle$ plus a change of dummy summation indexes in one of the two shows that this sum is of the form $Z + {{Z}}^{*}$. This sum is intrinsically real, so we can neglect one of the two doubling the other, but we will still be required to show that the real part of either is zero.
Lets reduce these one at a time starting with (12), and write $\dot{A}_\mathbf{k} = \kappa$ temporarily
\begin{aligned}\left\langle{{ (\gamma^0 \wedge {{\kappa}}^{*} ) (\gamma^0 \wedge \kappa }}\right\rangle &={\kappa^m}^{{*}} \kappa^{m'}\left\langle{{ \gamma^0 \gamma_m \gamma^0 \gamma_{m'} }}\right\rangle \\ &=-{\kappa^m}^{{*}} \kappa^{m'}\left\langle{{ \gamma_m \gamma_{m'} }}\right\rangle \\ &={\kappa^m}^{{*}} \kappa^{m'}\delta_{m m'}.\end{aligned}
So the first of our Hamiltonian terms is
\begin{aligned}\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_\mathbf{k}}}^{*} ) (\gamma^0 \wedge \dot{A}_\mathbf{k} }}\right\rangle &=\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2}{\left\lvert{{{\dot{A}}_{\mathbf{k}}}^m}\right\rvert}^2.\end{aligned} \quad\quad\quad(16)
Note that summation over $m$ is still implied here, so we’d be better off with a spatial vector representation of the Fourier coefficients $\mathbf{A}_\mathbf{k} = A_\mathbf{k} \wedge \gamma_0$. With such a notation, this contribution to the Hamiltonian is
\begin{aligned}\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2} \dot{\mathbf{A}}_\mathbf{k} \cdot {{\dot{\mathbf{A}}_\mathbf{k}}}^{*}.\end{aligned} \quad\quad\quad(17)
To reduce (13) and (13), this time writing $\kappa = A_\mathbf{k}$, we can start with just the scalar selection
\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) ( \gamma^0 \wedge \dot{\kappa} ) }}\right\rangle &=\Bigl( \gamma^m {{(\kappa^0)}}^{*} - {{\kappa}}^{*} \underbrace{(\gamma^m \cdot \gamma^0)}_{=0} \Bigr) \cdot \dot{\kappa} \\ &={{(\kappa^0)}}^{*} \dot{\kappa}^m\end{aligned}
Thus the contribution to the Hamiltonian from (13) and (13) is
\begin{aligned}\frac{2 \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \pi k_m}{c \lambda_m} \text{Real} \Bigl( i {{(A_\mathbf{k}^0)}}^{*} \dot{A_\mathbf{k}}^m \Bigl)=\frac{2 \pi \epsilon_0 \lambda_1 \lambda_2 \lambda_3}{c} \text{Real} \Bigl( i {{(A_\mathbf{k}^0)}}^{*} \mathbf{k} \cdot \dot{\mathbf{A}}_\mathbf{k} \Bigl).\end{aligned} \quad\quad\quad(18)
Most definitively not zero in general. Our final expansion (15) is the messiest. Again with $A_\mathbf{k} = \kappa$ for short, the grade selection of this term in coordinates is
\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) \gamma^0 (\gamma^n \wedge \kappa ) \gamma^0 }}\right\rangle&=- {{\kappa_\mu}}^{*} \kappa^\nu \left\langle{{ (\gamma^m \wedge \gamma^\mu) \gamma^0 (\gamma_n \wedge \gamma_\nu) \gamma^0 }}\right\rangle\end{aligned} \quad\quad\quad(19)
Expanding this out yields
\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) \gamma^0 (\gamma^n \wedge \kappa ) \gamma^0 }}\right\rangle&=- ( {\left\lvert{\kappa_0}\right\rvert}^2 - {\left\lvert{A^a}\right\rvert}^2 ) \delta_{m n} + {{A^n}}^{*} A^m.\end{aligned} \quad\quad\quad(20)
The contribution to the Hamiltonian from this, with $\phi_\mathbf{k} = A^0_\mathbf{k}$, is then
\begin{aligned}2 \pi^2 \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \Bigl(-\mathbf{k}^2 {{\phi_\mathbf{k}}}^{*} \phi_\mathbf{k} + \mathbf{k}^2 ({{\mathbf{A}_\mathbf{k}}}^{*} \cdot \mathbf{A}_\mathbf{k})+ (\mathbf{k} \cdot {{\mathbf{A}_k}}^{*}) (\mathbf{k} \cdot \mathbf{A}_k)\Bigr).\end{aligned} \quad\quad\quad(21)
A final reassembly of the Hamiltonian from the parts (17) and (18) and (21) is then
\begin{aligned}H = \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \sum_\mathbf{k}\left(\frac{1}{{2 c^2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{2 \pi}{c} \text{Real} \Bigl( i {{ \phi_\mathbf{k} }}^{*} (\mathbf{k} \cdot \dot{\mathbf{A}}_\mathbf{k}) \Bigl)+2 \pi^2 \Bigl(\mathbf{k}^2 ( -{\left\lvert{\phi_\mathbf{k}}\right\rvert}^2 + {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 ) + {\left\lvert{\mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(22)
This is finally reduced to a completely real expression, and one without any explicit Geometric Algebra. All the four vector Fourier vector potentials written out explicitly in terms of the spacetime split $A_\mathbf{k} = (\phi_\mathbf{k}, \mathbf{A}_\mathbf{k})$, which is natural since an explicit time and space split was the starting point.
# Gauge transformation to simplify the Hamiltonian.
While (22) has considerably simpler form than (11), what was expected, was something that looked like the Harmonic oscillator. On the surface this does not appear to be such a beast. Exploitation of gauge freedom is required to make the simplification that puts things into the Harmonic oscillator form.
If we are to change our four vector potential $A \rightarrow A + \nabla \psi$, then Maxwell’s equation takes the form
\begin{aligned}J/\epsilon_0 c = \nabla (\nabla \wedge (A + \nabla \psi) = \nabla (\nabla \wedge A) + \nabla (\underbrace{\nabla \wedge \nabla \psi}_{=0}),\end{aligned} \quad\quad\quad(23)
which is unchanged by the addition of the gradient to any original potential solution to the equation. In coordinates this is a transformation of the form
\begin{aligned}A^\mu \rightarrow A^\mu + \partial_\mu \psi,\end{aligned} \quad\quad\quad(24)
and we can use this to force any one of the potential coordinates to zero. For this problem, it appears that it is desirable to seek a $\psi$ such that $A^0 + \partial_0 \psi = 0$. That is
\begin{aligned}\sum_\mathbf{k} \phi_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} + \frac{1}{{c}} \partial_t \psi = 0.\end{aligned} \quad\quad\quad(25)
Or,
\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x},0) -\frac{1}{{c}} \sum_\mathbf{k} e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \int_{\tau=0}^t \phi_\mathbf{k}(\tau).\end{aligned} \quad\quad\quad(26)
With such a transformation, the $\phi_\mathbf{k}$ and $\dot{\mathbf{A}}_\mathbf{k}$ cross term in the Hamiltonian (22) vanishes, as does the $\phi_\mathbf{k}$ term in the four vector square of the last term, leaving just
\begin{aligned}H = \frac{\epsilon_0}{c^2} \lambda_1 \lambda_2 \lambda_3 \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} \Bigl((2 \pi c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 + {\left\lvert{ ( 2 \pi c \mathbf{k}) \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(27)
Additionally, wedging (5) with $\gamma_0$ now does not loose any information so our potential Fourier series is reduced to just
\begin{aligned}\mathbf{A} &= \sum_{\mathbf{k}} \mathbf{A}_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \\ \mathbf{A}_\mathbf{k} &= \frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} \mathbf{A}(\mathbf{x}, t) e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3.\end{aligned} \quad\quad\quad(28)
The desired harmonic oscillator form would be had in (27) if it were not for the $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ term. Does that vanish? Returning to Maxwell’s equation should answer that question, but first it has to be expressed in terms of the vector potential. While $\mathbf{A} = A \wedge \gamma_0$, the lack of an $A^0$ component means that this can be inverted as
\begin{aligned}A = \mathbf{A} \gamma_0 = -\gamma_0 \mathbf{A}.\end{aligned} \quad\quad\quad(30)
The gradient can also be factored scalar and spatial vector components | 95,443 | 261,417 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1020, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-10 | latest | en | 0.951928 |
https://www.stem.org.uk/resources/search?f%5B0%5D=field_age_range%3A22&f%5B1%5D=field_type%3A115&f%5B2%5D=field_age_range%3A96&%3Bf%5B1%5D=field_type%3A53&%3Bamp%3Bf%5B1%5D=field_age_range%3A83&%3Bamp%3Bf%5B2%5D=field_subject%3A20&%3Bamp%3Bf%5B3%5D=field_subject%3A26&%3Bamp%3Bf%5B4%5D=type%3Acommunity_resource&%3Bamp%3Bf%5B5%5D=field_age_range%3A3&%3Bamp%3Bf%5B6%5D=field_publication_year%3A84&%3Bamp%3Bf%5B7%5D=field_type%3A86 | 1,563,776,662,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527531.84/warc/CC-MAIN-20190722051628-20190722073628-00064.warc.gz | 856,791,203 | 12,152 | ## Listing all results (307)
### Ohm's Law
This simulation demonstrates how Ohm's law relates to a simple circuit. The voltage and resistance can be adjusted to see how this affects the current. The sizes of the symbols in the equation change to match the circuit diagram.
### Symmetry
In these AMP resources, co-funded by the Clothworkers and Nuffield Foundation, students investigate making different symmetrical shapes, using one or more of three given shapes. Students will work logically, identifying and classifying patterns making general statements.
The key processes are:
*...
### Water Availability
In this activity, students take the role of data analysts of the World Water Resources Board (WWRB) a fictional organisation charged with providing financial aid to countries most in need of water. They will compare the availability of water in Algeria, Jordan and Turkey. Students will determine how to fairly...
### Torbury Festival
Torbury Festival is a set of interactive lessons around staging of a music festival. To overcome various challenges, from floods to escaped cattle to over-excited crowds storming the stage, students must apply their mathematical knowledge to real life situations. The problems are intended to promote discussion,...
### Three Dice
In this Nuffield probability activity students play a bingo-style game. To maximize their chances of winning, students must decide which numbers are most likely to occur when three dice are thrown and the scores are added.
The key processes developed in this activity are:
*Representing - identifying...
### Stacks
In this Nuffield interactive activity students are asked to explore, analyse and describe the patterns generated by moving counters between two stacks according to a fixed rule, always doubling the size of the smaller stack.
The key processes developed by this activity include:
*Representing -...
This activity provides the opportunity for students to appreciate how mathematics plays a part in everyday life. The town council has allocated £100,000 to spend on reducing the number of deaths and serious injuries due to road accidents. Students are expected to adopt the role of planners who work in small teams...
### Point Zero
In this activity, students assume the roles of various characters placed in difficult situations e.g. investigating a malfunctioning lift system, attempting to avoid activation a floor-based alarm, escaping from an underground railway network. Students will need to use thinking skills, reasoning skills and problem...
### Outbreak
In this activity students explore the use of mathematics in a disease management context. Students assume the role of scientists attempting to contain the spread of a dangerous virus and are expected to locate infected people, develop a successful antidote and manage a large-scale vaccination programme. Students...
### Number Stairs
This Subtangent investigation looks at the total of the squares inside a stair shape.
Initially students investigate the relationship of the totals of a three step stair and its position on a 10 by 10 grid.
Using algebra, students may discover a general rule which applies to any stair on the grid.... | 597 | 3,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-30 | latest | en | 0.925106 |
http://en.wikipedia.org/wiki/Gibbs_phenomenon | 1,398,238,903,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00174-ip-10-147-4-33.ec2.internal.warc.gz | 122,838,873 | 22,053 | # Gibbs phenomenon
In mathematics, the Gibbs phenomenon, discovered by Henry Wilbraham (1848)[1] and rediscovered by J. Willard Gibbs (1899),[2] is the peculiar manner in which the Fourier series of a piecewise continuously differentiable periodic function behaves at a jump discontinuity: the nth partial sum of the Fourier series has large oscillations near the jump, which might increase the maximum of the partial sum above that of the function itself. The overshoot does not die out as the frequency increases, but approaches a finite limit.[3]
These are one cause of ringing artifacts in signal processing.
## Description
Functional approximation of square wave using 5 harmonics
Functional approximation of square wave using 25 harmonics
Functional approximation of square wave using 125 harmonics
The Gibbs phenomenon involves both the fact that Fourier sums overshoot at a jump discontinuity, and that this overshoot does not die out as the frequency increases.
The three pictures on the right demonstrate the phenomenon for a square wave (of height $\pi/4$) whose Fourier expansion is
$\sin(x)+\frac{1}{3}\sin(3x)+\frac{1}{5}\sin(5x)+\dotsb.$
More precisely, this is the function f which equals $\pi/4$ between $2n\pi$ and $(2n+1)\pi$ and $-\pi/4$ between $(2n+1)\pi$ and $(2n+2)\pi$ for every integer n; thus this square wave has a jump discontinuity of height $\pi/2$ at every integer multiple of $\pi$.
As can be seen, as the number of terms rises, the error of the approximation is reduced in width and energy, but converges to a fixed height. A calculation for the square wave (see Zygmund, chap. 8.5., or the computations at the end of this article) gives an explicit formula for the limit of the height of the error. It turns out that the Fourier series exceeds the height $\pi/4$ of the square wave by
$\frac{1}{2}\int_0^\pi \frac{\sin t}{t}\, dt - \frac{\pi}{4} = \frac{\pi}{2}\cdot (0.089490\dots)$
or about 9 percent. More generally, at any jump point of a piecewise continuously differentiable function with a jump of a, the nth partial Fourier series will (for n very large) overshoot this jump by approximately $a \cdot (0.089490\dots)$ at one end and undershoot it by the same amount at the other end; thus the "jump" in the partial Fourier series will be about 18% larger than the jump in the original function. At the location of the discontinuity itself, the partial Fourier series will converge to the midpoint of the jump (regardless of what the actual value of the original function is at this point). The quantity
$\int_0^\pi \frac{\sin t}{t}\ dt = (1.851937052\dots) = \frac{\pi}{2} + \pi \cdot (0.089490\dots)$
is sometimes known as the Wilbraham–Gibbs constant.
### History
The Gibbs phenomenon was first noticed and analyzed by the obscure Henry Wilbraham.[1] He published a paper on it in 1848 that went unnoticed by the mathematical world.[4] Albert Michelson developed a device in 1898 that could compute and re-synthesize the Fourier series. A widespread myth says that when the Fourier coefficients for a square wave were input to the machine, the graph would oscillate at the discontinuities, and that because it was a physical device subject to manufacturing flaws, Michelson was convinced that the overshoot was caused by errors in the machine. In fact the graphs produced by the machine were not good enough to exhibit the Gibbs phenomenon clearly, and Michelson may not have noticed it as he made no mention of this effect in his paper (Michelson & Stratton 1898) about his machine or his later letters to Nature. Inspired by some correspondence in Nature between Michelson and Love about the convergence of the Fourier series of the square wave function, in 1898 J. Willard Gibbs published a short note in which he considered what today would be called a sawtooth wave and pointed out the important distinction between the limit of the graphs of the partial sums of the Fourier series, and the graph of the function that is the limit of those partial sums. In his first letter Gibbs failed to notice the Gibbs phenomenon, and the limit that he described for the graphs of the partial sums was inaccurate. In 1899 he published a correction in which he described the overshoot at the point of discontinuity (Nature: April 27, 1899, p. 606). In 1906, Maxime Bôcher gave a detailed mathematical analysis of that overshoot, which he called the "Gibbs phenomenon".[5]
### Explanation
Informally, it reflects the difficulty inherent in approximating a discontinuous function by a finite series of continuous sine and cosine waves. It is important to put emphasis on the word finite because even though every partial sum of the Fourier series overshoots the function it is approximating, the limit of the partial sums does not. The value of x where the maximum overshoot is achieved moves closer and closer to the discontinuity as the number of terms summed increases so, again informally, once the overshoot has passed by a particular x, convergence at the value of x is possible.
There is no contradiction in the overshoot converging to a non-zero amount, but the limit of the partial sums having no overshoot, because where that overshoot happens moves. We have pointwise convergence, but not uniform convergence. For a piecewise C1 function the Fourier series converges to the function at every point except at the jump discontinuities. At the jump discontinuities themselves the limit will converge to the average of the values of the function on either side of the jump. This is a consequence of the Dirichlet theorem.[6]
The Gibbs phenomenon is also closely related to the principle that the decay of the Fourier coefficients of a function at infinity is controlled by the smoothness of that function; very smooth functions will have very rapidly decaying Fourier coefficients (resulting in the rapid convergence of the Fourier series), whereas discontinuous functions will have very slowly decaying Fourier coefficients (causing the Fourier series to converge very slowly). Note for instance that the Fourier coefficients 1, −1/3, 1/5, ... of the discontinuous square wave described above decay only as fast as the harmonic series, which is not absolutely convergent; indeed, the above Fourier series turns out to be only conditionally convergent for almost every value of x. This provides a partial explanation of the Gibbs phenomenon, since Fourier series with absolutely convergent Fourier coefficients would be uniformly convergent by the Weierstrass M-test and would thus be unable to exhibit the above oscillatory behavior. By the same token, it is impossible for a discontinuous function to have absolutely convergent Fourier coefficients, since the function would thus be the uniform limit of continuous functions and therefore be continuous, a contradiction. See more about absolute convergence of Fourier series.
### Solutions
In practice, the difficulties associated with the Gibbs phenomenon can be ameliorated by using a smoother method of Fourier series summation, such as Fejér summation or Riesz summation, or by using sigma-approximation. Using a wavelet transform with Haar basis functions, the Gibbs phenomenon does not occur in the case of continuous data at jump discontinuities,[7] and is minimal in the discrete case at large change points. In wavelet analysis, this is commonly referred to as the Longo phenomenon.
## Formal mathematical description of the phenomenon
Let $f: {\Bbb R} \to {\Bbb R}$ be a piecewise continuously differentiable function which is periodic with some period $L > 0$. Suppose that at some point $x_0$, the left limit $f(x_0^-)$ and right limit $f(x_0^+)$ of the function $f$ differ by a non-zero gap $a$:
$f(x_0^+) - f(x_0^-) = a \neq 0.$
For each positive integer N ≥ 1, let SN f be the Nth partial Fourier series
$S_N f(x) := \sum_{-N \leq n \leq N} \hat f(n) e^{\frac{2i\pi n x}{L}} = \frac{1}{2} a_0 + \sum_{n=1}^N \left( a_n \cos\left(\frac{2\pi nx}{L}\right) + b_n \sin\left(\frac{2\pi nx}{L}\right) \right),$
where the Fourier coefficients $\hat f(n), a_n, b_n$ are given by the usual formulae
$\hat f(n) := \frac{1}{L} \int_0^L f(x) e^{-2i\pi n x/L}\, dx$
$a_n := \frac{2}{L} \int_0^L f(x) \cos\left(\frac{2\pi nx}{L}\right)\, dx$
$b_n := \frac{2}{L} \int_0^L f(x) \sin\left(\frac{2\pi nx}{L}\right)\, dx.$
Then we have
$\lim_{N \to \infty} S_N f\left(x_0 + \frac{L}{2N}\right) = f(x_0^+) + a\cdot (0.089490\dots)$
and
$\lim_{N \to \infty} S_N f\left(x_0 - \frac{L}{2N}\right) = f(x_0^-) - a\cdot (0.089490\dots)$
but
$\lim_{N \to \infty} S_N f(x_0) = \frac{f(x_0^-) + f(x_0^+)}{2}.$
More generally, if $x_N$ is any sequence of real numbers which converges to $x_0$ as $N \to \infty$, and if the gap a is positive then
$\limsup_{N \to \infty} S_N f(x_N) \leq f(x_0^+) + a\cdot (0.089490\dots)$
and
$\liminf_{N \to \infty} S_N f(x_N) \geq f(x_0^-) - a\cdot (0.089490\dots).$
If instead the gap a is negative, one needs to interchange limit superior with limit inferior, and also interchange the ≤ and ≥ signs, in the above two inequalities.
## Signal processing explanation
The sinc function, the impulse response of an ideal low-pass filter. Scaling narrows the function, and correspondingly increases magnitude (which is not shown here), but does not reduce the magnitude of the undershoot, which is the integral of the tail.
From the point of view of signal processing, the Gibbs phenomenon is the step response of a low-pass filter, and the oscillations are called ringing or ringing artifacts. Truncating the Fourier transform of a signal on the real line, or the Fourier series of a periodic signal (equivalently, a signal on the circle) corresponds to filtering out the higher frequencies by an ideal (brick-wall) low-pass/high-cut filter. This can be represented as convolution of the original signal with the impulse response of the filter (also known as the kernel), which is the sinc function. Thus the Gibbs phenomenon can be seen as the result of convolving a Heaviside step function (if periodicity is not required) or a square wave (if periodic) with a sinc function: the oscillations in the sinc function cause the ripples in the output.
The sine integral, exhibiting the Gibbs phenomenon for a step function on the real line.
In the case of convolving with a Heaviside step function, the resulting function is exactly the integral of the sinc function, the sine integral; for a square wave the description is not as simply stated. For the step function, the magnitude of the undershoot is thus exactly the integral of the (left) tail, integrating to the first negative zero: for the normalized sinc of unit sampling period, this is $\int_{-\infty}^{-1} \frac{\sin(\pi x)}{\pi x}\,dx.$ The overshoot is accordingly of the same magnitude: the integral of the right tail, or, which amounts to the same thing, the difference between the integral from negative infinity to the first positive zero, minus 1 (the non-overshooting value).
The overshoot and undershoot can be understood thus: kernels are generally normalized to have integral 1, so they result in a mapping of constant functions to constant functions – otherwise they have gain. The value of a convolution at a point is a linear combination of the input signal, with coefficients (weights) the values of the kernel. If a kernel is non-negative, such as for a Gaussian kernel, then the value of the filtered signal will be a convex combination of the input values (the coefficients (the kernel) integrate to 1, and are non-negative), and will thus fall between the minimum and maximum of the input signal – it will not undershoot or overshoot. If, on the other hand, the kernel assumes negative values, such as the sinc function, then the value of the filtered signal will instead be an affine combination of the input values, and may fall outside of the minimum and maximum of the input signal, resulting in undershoot and overshoot, as in the Gibbs phenomenon.
Taking a longer expansion – cutting at a higher frequency – corresponds in the frequency domain to widening the brick-wall, which in the time domain corresponds to narrowing the sinc function and increasing its height by the same factor, leaving the integrals between corresponding points unchanged. This is a general feature of the Fourier transform: widening in one domain corresponds to narrowing and increasing height in the other. This results in the oscillations in sinc being narrower and taller and, in the filtered function (after convolution), yields oscillations that are narrower and thus have less area, but does not reduce the magnitude: cutting off at any finite frequency results in a sinc function, however narrow, with the same tail integrals. This explains the persistence of the overshoot and undershoot.
Thus the features of the Gibbs phenomenon are interpreted as follows:
• the undershoot is due to the impulse response having a negative tail integral, which is possible because the function takes negative values;
• the overshoot offsets this, by symmetry (the overall integral does not change under filtering);
• the persistence of the oscillations is because increasing the cutoff narrows the impulse response, but does not reduce its integral – the oscillations thus move towards the discontinuity, but do not decrease in magnitude.
## The square wave example
Animation of the additive synthesis of a square wave with an increasing number of harmonics. The Gibbs phenomenon is visible especially when the number of harmonics is large.
We now illustrate the above Gibbs phenomenon in the case of the square wave described earlier. In this case the period L is $2\pi$, the discontinuity $x_0$ is at zero, and the jump a is equal to $\pi/2$. For simplicity let us just deal with the case when N is even (the case of odd N is very similar). Then we have
$S_N f(x) = \sin(x) + \frac{1}{3} \sin(3x) + \cdots + \frac{1}{N-1} \sin((N-1)x).$
Substituting $x=0$, we obtain
$S_N f(0) = 0 = \frac{-\frac{\pi}{4} + \frac{\pi}{4}}{2} = \frac{f(0^-) + f(0^+)}{2}$
as claimed above. Next, we compute
$S_N f\left(\frac{2\pi}{2N}\right) = \sin\left(\frac{\pi}{N}\right) + \frac{1}{3} \sin\left(\frac{3\pi}{N}\right) + \cdots + \frac{1}{N-1} \sin\left( \frac{(N-1)\pi}{N} \right).$
If we introduce the normalized sinc function, $\operatorname{sinc}(x)\,$, we can rewrite this as
$S_N f\left(\frac{2\pi}{2N}\right) = \frac{\pi}{2} \left[ \frac{2}{N} \operatorname{sinc}\left(\frac{1}{N}\right) + \frac{2}{N} \operatorname{sinc}\left(\frac{3}{N}\right) + \cdots + \frac{2}{N} \operatorname{sinc}\left( \frac{(N-1)}{N} \right) \right].$
But the expression in square brackets is a numerical integration approximation to the integral $\int_0^1 \operatorname{sinc}(x)\ dx$ (more precisely, it is a midpoint rule approximation with spacing $2/N$). Since the sinc function is continuous, this approximation converges to the actual integral as $N \to \infty$. Thus we have
\begin{align} \lim_{N \to \infty} S_N f\left(\frac{2\pi}{2N}\right) & = \frac{\pi}{2} \int_0^1 \operatorname{sinc}(x)\, dx \\[8pt] & = \frac{1}{2} \int_{x=0}^1 \frac{\sin(\pi x)}{\pi x}\, d(\pi x) \\[8pt] & = \frac{1}{2} \int_0^\pi \frac{\sin(t)}{t}\ dt \quad = \quad \frac{\pi}{4} + \frac{\pi}{2} \cdot (0.089490\dots), \end{align}
which was what was claimed in the previous section. A similar computation shows
$\lim_{N \to \infty} S_N f\left(-\frac{2\pi}{2N}\right) = -\frac{\pi}{2} \int_0^1 \operatorname{sinc}(x)\ dx = -\frac{\pi}{4} - \frac{\pi}{2} \cdot (0.089490\dots).$
## Consequences
In signal processing, the Gibbs phenomenon is undesirable because it causes artifacts, namely clipping from the overshoot and undershoot, and ringing artifacts from the oscillations. In the case of low-pass filtering, these can be reduced or eliminated by using different low-pass filters.
In MRI, the Gibbs phenomenon causes artifacts in the presence of adjacent regions of markedly differing signal intensity. This is most commonly encountered in spinal MR imaging, where the Gibbs phenomenon may simulate the appearance of syringomyelia. | 4,065 | 16,114 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 50, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2014-15 | longest | en | 0.872496 |
http://www.physicsforums.com/showthread.php?t=38391 | 1,398,403,649,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00298-ip-10-147-4-33.ec2.internal.warc.gz | 805,555,851 | 10,405 | # Solving for the Determinent of a Matrix
by Poweranimals
Tags: determinent, matrix, solving
P: 68 Okay, I'm learning currently how to solve for the determinent of a Matrix. Of course the book explains how to solve for a 2 X 2 Matrix, a 3 X 3 Matrix, a 4 X 4 Matrix, ect. But it says nothing about how to solve for a 3 X 2 Matrix. Any idea how to do this? I'm really baffled on this.
P: 68 Just in case no one knows what I'm talking about, by a 3 X 2 Matrix, I mean like this: [1 0 -2] [2 1 -1] It's very hard to figure this out. My book doesn't even go over it. They'll go over 2 X 2, 3 X 3, 4 X 4, 5 X 5, 6 X 6, ect.. But for some reason they don't touch on if the Matrix doesn't have equal sides.
P: 696
But it says nothing about how to solve for a 3 X 2 Matrix.
No wonder, the determinant function (or "a determinantal function") is defined as a function from the set of all nxn (i.e square) matrices (with elements in a field F), to the field F (the determinant takes a square matrix and spits back out a number). There are reasons for this.
"The" inverse of a matrix A is a matrix B such that AB = BA = I (i.e. B is both a left inverse and a right inverse of A). Suppose A is of size nxm, and B is pxq. Then AB is nxq and BA is pxm. But I is square, say I is a txt matrix. Since AB = BA = I, this forces n = q = p = m = t, i.e. A and B are both square. Thus only square matrices can have inverses.
One wants the determinant function to characterize when exactly a matrix X has an inverse (it just so happens to be that X has an inverse iff det(X) != 0). But according to the above, only square matrices can have inverses, so it doesn't make much sense to define the determinant for non-square matrices.
Btw, a 3x2 matrix has 3 rows and 2 columns. The matrix you used as an example is a 2x3 matrix.
P: 68
## Solving for the Determinent of a Matrix
So is it possible to solve the determinent of non square Matrices? Or is the answer simply Cannot be calculated?
P: 696 No, using the ordinary definition of the determinant function, you cannot calculate the determinant of a non-square matrix.
Math Emeritus Sci Advisor Thanks PF Gold P: 38,904 It's not so much that you "cannot calculate" it as that it is not defined! The determinant is only defined for square matrices. Asking how to find the determinant of a 3 by 2 matrix is a lot like asking how to find the square root of a chair.
P: 1 you can solve a 2x3 matrix, you use this all the time to when multiplying vectors. [a b c] [e f g] (bg-cf)-(ag-ce)+(af-be)
HW Helper
P: 3,225
Quote by korciuch you can solve a 2x3 matrix, you use this all the time to when multiplying vectors. [a b c] [e f g] (bg-cf)-(ag-ce)+(af-be)
So, this is called the korciuch-operator?
Math Emeritus Sci Advisor Thanks PF Gold P: 38,904 The phrase "solve a matrix" doesn't make sense. The orginal question was "solve for the determinant of a matrix". Again, a 3 by 2 matrix (or any non-square matrix) does not have a determinant.
P: 356
Quote by korciuch you can solve a 2x3 matrix, you use this all the time to when multiplying vectors. [a b c] [e f g] (bg-cf)-(ag-ce)+(af-be)
thats the cross product, which in this case, is a 3x3 matrix with [i j k] as the first row:
[i j k]
[a b c]
[e f g]
Sci Advisor HW Helper P: 9,428 the det of a non square matrix is either not defined or zero.
Math Emeritus Sci Advisor Thanks PF Gold P: 38,904 Could you give an example of a non-square matrix that has determanant 0?
P: 406 You could try the products of the nonzero singular values in the singular value decomposition. It's not the determinant, but it's the closest thing you're going to get.
Math
Emeritus | 1,028 | 3,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2014-15 | latest | en | 0.938103 |
https://goprep.co/ex-18.2-q7-the-breadth-of-a-room-is-twice-its-height-one-i-1njo4r | 1,618,136,554,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061820.19/warc/CC-MAIN-20210411085610-20210411115610-00246.warc.gz | 393,878,615 | 34,006 | Given,
Breadth of room = 2×height of room = b = 2×h = h =
Volume of room = 512 dm3
= l×b×h = 512
= 2b×b×
= b3 = 512 = b =
So, breadth of room = 8 dm
Length of room = 2b = 2×8 = 16 dm
Height of room =
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View Full Version : How to get the distance of camera from objects plane?
rakeshthp
09-26-2013, 12:02 PM
Hello,
I am working on rendering vector plot. I have the following projection settings
glMatrixMode(GL_PROJECTION);
glOrtho(centerp.x-screenwidth/2, centerp.x+screenwidth/2,
centerp.y-screenheight/2, centerp.y+screenheight/2, 1.0, 10000.0);
With the above projection, how can I calculate the distance of the object's plane to the camera's plane. i.e. how can I find from how much distance, I am viewing the object?
I need this, in order to perform scaling of vector arrows(glyphs). i.e. as I zoom in/out, the vector arrows should be scaled in order to maintain its actual dimension. So I thought probably based on the distance of camera from the object plane, the vector arrows can be scaled.
Thanks & Regards
Rakesh Patil
GClements
09-27-2013, 08:46 AM
I am working on rendering vector plot. I have the following projection settings
glOrtho(centerp.x-screenwidth/2, centerp.x+screenwidth/2,
centerp.y-screenheight/2, centerp.y+screenheight/2, 1.0, 10000.0);
With the above projection, how can I calculate the distance of the object's plane to the camera's plane. i.e. how can I find from how much distance, I am viewing the object?
That question isn't meaningful. An orthographic projection has the viewpoint infinitely far away from any object.
i.e. as I zoom in/out, the vector arrows should be scaled in order to maintain its actual dimension.
You can't "zoom" an orthographic projection; the field-of-view angle is always zero.
Carmine
09-27-2013, 09:41 AM
glMatrixMode(GL_PROJECTION);
glOrtho(centerp.x-screenwidth/2, centerp.x+screenwidth/2,
centerp.y-screenheight/2, centerp.y+screenheight/2, 1.0, 10000.0);
With the above projection, how can I calculate the distance of the object's plane to the camera's plane. i.e. how can I find from how much distance, I am viewing the object? I need this, in order to perform scaling of vector arrows(glyphs). i.e. as I zoom in/out, the vector arrows should be scaled in order to maintain its actual dimension. So I thought probably based on the distance of camera from the object plane, the vector arrows can be scaled. Not really sure what you are asking. Do you want the arrows to change size as you zoom in or out? Or, do you want them to remain the same size as you zoom in and out? Are your arrows raster patterns (like raster font), or are they OpenGL lines, polygons, etc.
rakeshthp
09-29-2013, 09:15 PM
@GClements:
I perform zooming by changing the values of left, right, top and bottom. I perform panning also by changing the left, right, top and bottom parameters. I don't know whether it is correct way of doing, but it does work for 2D data. If this is not the correct approach, then kindly guide me how to get correct zooming and panning.
@Carmine:
My arrows are opengl lines only currently. To answer your first question, suppose I am viewing at four cars from a distance 'd' where all four cars appear to be of actual size. Now if I move away from those cars, in real world, the cars would appear small sized. And if I move near to them they would appear much bigger, or in some case, some part of car would be out of my vision if I go more closer. I don't want this effect with the arrows. I want the arrows to look of same size, irrelevant of the distance of camera. So, it is quite obvious that as the camera moves away from arrows, the arrows must be scaled with large value to get the same size, and the arrows must be scaled with small value as the camera moves towards the arrow.
I hope I've made it more clearer as what I want to achieve.
Thanks & Regards
Rakesh Patil
wmchiew
09-29-2013, 10:58 PM
Hi Rakesh, since you are using glOrtho you are doing parallel projection in your scene. Thus your objects (the arrows) will not change in size on your viewing plane regardless of the distance of the camera. I believe this is what GClements is saying :)
rakeshthp
09-30-2013, 09:48 AM
Thus your objects (the arrows) will not change in size on your viewing plane regardless of the distance of the camera
Yes. I agree, and that's why I want to scale the arrows as the camera moves front and back.
Thanks & regards
Rakesh Patil
Carmine
09-30-2013, 10:18 AM
My arrows are opengl lines only currently. To answer your first question, suppose I am viewing at four cars from a distance 'd' where all four cars appear to be of actual size. Now if I move away from those cars, in real world, the cars would appear small sized. And if I move near to them they would appear much bigger, or in some case, some part of car would be out of my vision if I go more closer. I don't want this effect with the arrows. I want the arrows to look of same size, irrelevant of the distance of camera. So, it is quite obvious that as the camera moves away from arrows, the arrows must be scaled with large value to get the same size, and the arrows must be scaled with small value as the camera moves towards the arrow. So your are zooming in and out by changing the width (w) and height (h) in your glOrtho command. There is really no distance from the camera to your objects in an orthographic projection, which is why you can't figure out what 'd' is. What should be done is to scale each car around its centroid by the inverse of the relative change in w and/or h. For example, say you double the width of your window but don't change the height. Each car should be scaled by 0.5 in the x direction and no scaling in y. The GL command would be glScalef (0.5, 1.0, 1.0), where 0.5 is the inverse of the relative change in viewport width. A more detailed example: if the original window was 500 units wide and 300 units high, and the new window is 700 units wide by 400 units high, the gl command would be glScalef (sx, sy, 1), where sx = 500/700, and sy = 300/400. The scale command must be applied to each car individually about its centroid. This means: translate the car to the origin, apply the scaling, then transfer the car back to where it started. Give this a try and let us know how it goes.
Roaoul
09-30-2013, 02:26 PM
glOrtho() is good for taking measurements off the screen, but not for undistorted viewing with the screen at arms length, or with a large screen in a small room. For that you should use gluPerspective() if viewing on-axis, or glFrustum() if viewing off-axis.
To do what you want with glOrtho, you just need to keep track of the scaling and rotating you are doing (in eye coordinates) in your projection matrix with glOrtho, and undo that scaling and rotation in Object coordinates for your arrow: Place the head of your arrow at (0,0,0) in arrow object coordinates. Place the tail some standard distance away in +x. Then do the following matrix multiplies on the arrow-side (right side) of the m-v-p transformation, 1) inverse-glOrtho-scale, 2) inverse-glOrtho-rotate 3) translate in car coordinates to place on the car, then multiply on the left by the model-view-projection matrix you use to view the car. I can be more explicit if you need.
rakeshthp
10-01-2013, 09:16 PM
@Roaoul
Are you trying to explain the same thing what carmine explained?
Roaoul
10-02-2013, 12:56 PM
@Roaoul
Are you trying to explain the same thing what carmine explained?
Not much. I think we have different ideas of what you are trying to do. Carmine's directions would replace calling glOrtho() on resizing/shaping your window, I believe. If you follow those directions to replace the projection matrix, and you also separately scale your cars as directed, then I think you'll get a doubling of the scaling of the cars, not a zeroing of scaling of arrows.
But I interpreted your post as that you are trying to zoom in and out of some cars, using glOrtho() on the projection matrix, and want some annotation arrows you are drawing pointing to the cars to remain the same size on screen regardless of whether the cars are tiny or huge. Right?
My earlier post mistakenly included rotation considerations. You have none in the projection matrix to worry about, so please disregard all my rotation remarks (step 2). The rest is good. But you know, if you rotate your view (with modelView), then the arrows will also rotate unless you undo that too.
So anyway, given a beginning r,l,t,b,n,f inputs for your reference glOrtho() projection matrix, here is how to zoom and leave your arrows unchanged in size on screen:
When you zoom, you must change l-r and t-b and n-f all by the same factor, say zf. Zoom out is zf > 1; zoom in is zf < 1. In your code above, you would multiply screenwidth and screenheight by *zf. Also make sure your new n-f ends up zf*(n_old-f_old), or your near and far clipping planes won't zoom in and out (and you might get artifacts if you rotate your view).
Place your camera with the model-view matrix, and push it before placing each car. Place the car with glTranslate on the modelview matrix. Render the car and push the matrix for the arrows. For each arrow on that car, call glTranslate for where on the car you want the arrow to point, glScale(1/zf, 1/zf, 1/zf) (undoes your zoom), render your arrow, and pop for the next arrow. After the arrows, pop for the next car. I hope that helps!
rakeshthp
10-02-2013, 06:49 PM
@Roaoul
Thanks for the explaination. Firstly, I am sorry for improperly presenting what I require. The car's post was just an example. What I actually need is to render vector plot (quiver plot). The car has nothing to do with the arrows. I am working on a scientific application where I need to display vectors over a mesh. Like this one here (http://www.google.co.in/imgres?sa=X&es_sm=93&biw=1517&bih=725&tbm=isch&tbnid=Esig2NlzW6QxxM:&imgrefurl=http://orion.math.iastate.edu/reu/2001/matlab_graphics/matlab_graphics.html&docid=KG6ysBPKyzVDIM&imgurl=http://orion.math.iastate.edu/reu/2001/matlab_graphics/plot17.jpg&w=1200&h=900&ei=ZcxMUtr_LdCmrAf4s4GwAg&zoom=1&ved=1t:3588,r:1,s:0,i:81&iact=rc&page=1&tbnh=186&tbnw=259&start=0&ndsp=18&tx=145&ty=62). The concept of arrows what you interpreted is right. Regardless of what amount user zoom's in/out, the size of arrows should appear same for the user.
Let me explain why I need to scale the arrow size. Suppose the mesh is a huge mesh and the mesh nodes are very densely placed (i.e. very close to each other node). In this case I would get vector plot also dense. i.e. a black patch would appear at that region. So as a user, I would like to zoom that area and see what exactly is happening in that area. Now when I zoom, if I do not scale arrows, they will appear as a patch only. If I scale those arrows, then user will get clear idea what is going on. That's the main Idea behind it.
Anyways, I shall try what you told and get back to you. If you have any further clarification of suggestions, kindly post it here.
Thanks & Regards
Rakesh Patil
Roaoul
10-02-2013, 07:44 PM
@rakeshthp
How cool. I did something similar for my dissertation, but did it all in PostScript written by FORTRAN, since OpenGL wasn't available way back then. I still have the code. Is your mesh flat like a desktop or curved like a ball? What I gave you should work for flat, but you would probably want more for a curved surface (various rotation features).
Regards, Paul
rakeshthp
10-03-2013, 05:48 AM
@rakeshthp
How cool. I did something similar for my dissertation, but did it all in PostScript written by FORTRAN, since OpenGL wasn't available way back then. I still have the code. Is your mesh flat like a desktop or curved like a ball? What I gave you should work for flat, but you would probably want more for a curved surface (various rotation features).
Regards, Paul
Thanks a lot Paul. That really helped me a lot. Well, I would need your advice later stage also. Because presently this is for a two dimensional grid in a single plane (flat). For 3d grid whether it will be same technique, or it would change?
Thanks & Regards
Rakesh Patil | 3,063 | 11,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-30 | latest | en | 0.863724 |
https://homework.cpm.org/category/CCI_CT/textbook/Calc3rd/chapter/Ch9/lesson/9.3.1/problem/9-85 | 1,603,248,209,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874637.23/warc/CC-MAIN-20201021010156-20201021040156-00575.warc.gz | 365,594,550 | 15,112 | ### Home > CALC3RD > Chapter Ch9 > Lesson 9.3.1 > Problem9-85
9-85.
Convert each of the following sets of parametric equations to rectangular form.
1. $x = \sin^2(t)$ and $y =\sqrt [ 3 ] { \operatorname { sin } ( t ) }$
$(\sin^{1/3}(t))^6=\sin^2(t)$
1. $x = t^{12}$
$(t^4)^3=t^{12}$ and $y = t^4$ | 125 | 303 | {"found_math": true, "script_math_tex": 6, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2020-45 | latest | en | 0.44363 |
http://math.stackexchange.com/questions/314131/picture-of-a-4d-knot/318377 | 1,455,021,074,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701157075.54/warc/CC-MAIN-20160205193917-00122-ip-10-236-182-209.ec2.internal.warc.gz | 152,205,369 | 19,632 | # Picture of a 4D knot
A knot is a way to put a circle into 3-space $S^1 \to \mathbb R^3$ and these are often visualized as 2D knot diagrams.
Can anyone show me a diagram of a nontrivial knotted sphere $S^2 \to \mathbb R^4$ (differentiable)?
-
does the klien bottle count? I don't think its one but I am not sure. if it is I would like a different one. – user58512 Feb 25 '13 at 18:21
The Klein bottle does not count as a 'knotted sphere' - it's not a sphere in the first place! (That is, there's no diffeomorphism $S^2\to K$, because any such would preserve orientation and while $S^2$ is orientable, $K$ isn't.) It's arguably much closer to a torus than a sphere, anyway... – Steven Stadnicki Feb 25 '13 at 18:23
Have you seen this or this? – dtldarek Mar 2 '13 at 10:35
Here is a picture of a 4D knot. Most of the picture is embedded in a $3D$ time slice at time $t=0$. Then as you let $t$ increase the two top boundary circles persist, until you reach $t=1$ when they are capped with a disk. Similarly you cap off the two bottom boundary components with disks as $t$ decreases.
You can see that this construction yields a $2$-sphere since it represents two cylinders joined by a tube with caps added on the cylinder's ends.
-
Very nice picture! I can see that this really is a picture of a 2-sphere, but what kind of invariants would be used to show this 2-sphere is actually knotted? – Jason DeVito Mar 8 '13 at 16:29
@JasonDeVito: The fundamental group of the complement should do the trick, which you can compute using Seifert-van Kampen. – Grumpy Parsnip Mar 9 '13 at 15:20
The book How Surfaces Intersect in Space by J. Scott Carter covers this subject well, starting on page 226. You can get a free PDF of this book at http://www.maths.ed.ac.uk/~aar/papers/cartert.pdf . The book Surfaces in 4-Space by Carter, Kamata, and Saito covers this in-depth.
The premise of these books is that surfaces in 4-space can be represented as movies of embedded curves in 3-space, knotted or not.
If you want to see such a diagram Right Now®, then here is the first such diagram from Surfaces in 4-Space, the Fox-Milnor knotted sphere, which I am posting here FOR EDUCATIONAL PURPOSES ONLY, NOT FOR PROFIT:
-
Thank you very much! – user58512 Mar 2 '13 at 0:37
Oh man, for the longest time I was staring at the transition from $t=3$ to $t=4$ and seeing nothing more than this. If anyone else is as confused as I was, all that's happening is that they take the two loops and just wind them around each other. – Rahul Mar 2 '13 at 3:07
LOL! No, literally, I mean it. – Robert Haraway Mar 2 '13 at 4:22
+1 for the link. Carter's book looks very good. I'm hooked at the preface: "Milnor's book Topology from the Differentiable Viewpoint is a model for clarity, conciseness, and rigor. The current text might be subtitled Topology from Scott Carter's Viewpoint, and critics will, no doubt, say it is a model for none of the above." :-) – Jesse Madnick Mar 2 '13 at 9:50
Ralph Fox's "A Quick Trip Through Knot Theory" [1] has a lot of examples in section 6.
1. In Topology of 3-Manifolds and Related Topics, M.K. Fort (ed.), 1962.
- | 899 | 3,132 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2016-07 | longest | en | 0.939249 |
https://www.javatpoint.com/bfs-code-in-cpp | 1,720,978,937,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00238.warc.gz | 720,380,328 | 24,430 | # BFS Code in C++
## What is BFS?
Breadth-First Search (BFS) is an algorithm for traversing or searching a graph. It starts at a given vertex and explores all the neighboring vertices before moving on to the next level of vertices. BFS is useful for finding the shortest path between two vertices in a graph or for finding the connected components of a graph. It is also used for topological sorting, a linear ordering of the vertices in a directed acyclic graph (DAG) such that for every edge UV from vertex u to vertex v, u comes before v in the ordering.
In this tutorial, we will learn how to implement BFS in C++. We will start by understanding the concept of BFS and its time complexity. Then, we will see how to represent a graph in C++ using an adjacency list and how to traverse the graph using the BFS algorithm.
### Concept of BFS
BFS is an algorithm for traversing or searching a graph. It starts at a given vertex and explores all the neighboring vertices before moving on to the next level of vertices. The algorithm maintains a queue of vertices to be visited, starting with the source vertex. At each step, it removes a vertex from the queue and adds its unvisited neighbors to the queue. This process continues until no more vertices are in the queue to be visited.
The BFS algorithm has a time complexity of O(V+E), where V is the number of vertices and E is the number of edges in the graph. This makes it more suitable for dense graphs, where the number of edges is close to the number of vertices.
### Representing a Graph in C++
There are several ways to represent a graph in C++. One common representation is to use an adjacency list, which is a list of all the vertices that are adjacent to a given vertex. For example, consider the following graph with 5 vertices:
We can represent this graph using an adjacency list as follows:
Here, adj_list[i] is a vector containing the neighbors of vertex i.
Implementing BFS in C++
Output:
Explanation:
In this code, we first define the bfs() function, which takes an adjacency list adj_list, a starting vertex start, and an optional target vertex target. The bfs() function initializes a boolean array visited to track the visited vertices and a vector order to store the BFS traversal order. It also creates a queue q to hold the vertices to be visited.
The function then adds the starting vertex to the queue and marks it as visited. It then enters a loop that continues until the queue is empty. Each iteration of the loop removes the next vertex from the queue and adds it to the traversal order. It then adds all its unvisited neighbors to the queue and marks them as visited.
Finally, if a target vertex is specified and it is not found, the function returns an empty vector. Otherwise, it returns the traversal order.
In the main() function, we create an adjacency list for a graph with 5 vertices. We then perform a BFS search starting from vertex 0 and store the traversal order in the order vector. Finally, we print the traversal order to the console. | 638 | 3,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-30 | latest | en | 0.929735 |
http://www.studymode.com/essays/Some-Application-Of-Calculus-664406.html | 1,490,664,127,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189583.91/warc/CC-MAIN-20170322212949-00082-ip-10-233-31-227.ec2.internal.warc.gz | 698,277,225 | 19,171 | # Some Application of Calculus
Only available on StudyMode
• Topic: Derivative, Calculus, Differential calculus
• Pages : 5 (1446 words )
• Published : April 16, 2011
Text Preview
Calculus:
Calculus (Latin, calculus, a small stone used for counting) is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education. It has two major branches, Differential Calculus and Integral Calculus, which are related by the fundamental theorem of calculus.
Calculus is the study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations. A course in calculus is a gateway to other, more advanced courses in mathematics devoted to the study of functions and limits, broadly called mathematical analysis. Calculus has widespread applications in science, economics, and engineering and can solve many problems for which algebra alone is insufficient.
Application of Calculus in Various Fields:
Calculus is used in every branch of the physical sciences, actuarial science, computer science, statistics, engineering, economics, business, medicine, demography, and in other fields wherever a problem can be mathematically modeled and an optimal solution is desired. It allows one to go from (non-constant) rates of change to the total change or vice versa, and many times in studying a problem we know one and are trying to find the other.
Physics makes particular use of calculus; all concepts in classical mechanics and electromagnetism are interrelated through calculus. The mass of an object of known density, the moment of inertia of objects, as well as the total energy of an object within a conservative field can be found by the use of calculus. An example of the use of calculus in mechanics is Newton's second law of motion: historically stated it expressly uses the term "rate of change" which refers to the derivative saying the rate of change of momentum of a body is equal to the resultant force acting on the body and is in the same direction. Commonly expressed today as Force = Mass × acceleration, it involves differential calculus because acceleration is the time derivative of velocity or second time derivative of trajectory or spatial position. Starting from knowing how an object is accelerating, we use calculus to derive its path.
Maxwell's theory of electromagnetism and Einstein's theory of general relativity are also expressed in the language of differential calculus. Chemistry also uses calculus in determining reaction rates and radioactive decay. In biology, population dynamics starts with reproduction and death rates to model population changes.
Calculus can be used in conjunction with other mathematical disciplines. For example, it can be used with linear algebra to find the "best fit" linear approximation for a set of points in a domain. Or it can be used in probability theory to determine the probability of a continuous random variable from an assumed density function. In analytic geometry, the study of graphs of functions, calculus is used to find high points and low points (maxima and minima), slope, concavity and inflection points.
Green's Theorem, which gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C, is applied in an instrument known as a planimeter which is used to calculate the area of a flat surface on a drawing. For example, it can be used to calculate the amount of area taken up by an irregularly shaped flower bed or swimming pool when designing the layout of a piece of property.
In the realm of medicine, calculus can be used to find the optimal branching angle of a blood vessel so as to maximize flow. From the decay laws for a particular drug's elimination from the body, it's used to derive dosing laws. In nuclear medicine, it's used to build models of radiation transport in targeted tumor therapies. | 794 | 4,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-13 | latest | en | 0.936905 |
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A185903 T(n,k)=1/8 the number of nXk 0..7 arrays with every element equal to exactly one or two of its horizontal and vertical neighbors 4
0, 1, 1, 1, 15, 1, 8, 196, 196, 8, 15, 2765, 7854, 2765, 15, 71, 38731, 546280, 546280, 38731, 71, 176, 545328, 28967421, 139230658, 28967421, 545328, 176, 673, 7675381, 1748982326, 32611077940, 32611077940, 1748982326, 7675381, 673, 1905 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,5 COMMENTS Table starts ....0.........1...........1.............8.............15............71 ....1........15.........196..........2765..........38731........545328 ....1.......196........7854........546280.......28967421....1748982326 ....8......2765......546280.....139230658....32611077940.7921723799549 ...15.....38731....28967421...32611077940.30884279980274 ...71....545328..1748982326.7921723799549 ..176...7675381.98953708266 ..673.108065699 .1905 LINKS EXAMPLE Some solutions for 4X3 with a(1,1)=0 ..0..4..5....0..0..4....0..4..0....0..0..6....0..4..2....0..4..6....0..0..3 ..0..4..5....4..4..4....0..4..0....4..4..6....0..4..2....0..4..6....4..0..3 ..1..1..4....6..7..3....6..6..0....4..4..5....5..3..3....2..0..0....4..0..0 ..1..1..4....6..7..3....5..5..0....3..3..5....5..5..5....2..5..5....4..4..4 CROSSREFS Column 1 is A015442(n-1) Sequence in context: A040234 A040235 A040236 * A040237 A317835 A225762 Adjacent sequences: A185900 A185901 A185902 * A185904 A185905 A185906 KEYWORD nonn,tabl AUTHOR R. H. Hardin Feb 06 2011 STATUS approved
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Last modified March 21 22:19 EDT 2019. Contains 321382 sequences. (Running on oeis4.) | 723 | 1,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-13 | latest | en | 0.534422 |
https://www.jamiletheteacher.com/geometry/question-how-hard-is-geometry.html | 1,660,342,711,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00039.warc.gz | 749,329,344 | 11,143 | ## Question: How Hard Is Geometry?
Why is geometry difficult? Geometry is creative rather than analytical, and students often have trouble making the leap between Algebra and Geometry. They are required to use their spatial and logical skills instead of the analytical skills they were accustomed to using in Algebra.
## Is geometry hard in high school?
It is not any secret that high school geometry with its formal (two-column) proofs is considered hard and very detached from practical life. Many teachers in public school have tried different teaching methods and programs to make students understand this formal geometry, sometimes with success and sometimes not.
## What is the hardest thing in geometry?
The problem is known as Langley’s Adventitious Angles and was posed in 1922. It is also known as the hardest easy geometry problem because it can be solved by elementary methods but it is difficult and laborious.
## Do most people fail geometry?
According to the most recent data, 71 percent of high school students failed geometry exams in June, and 68 percent flunked Algebra 1 finals. Exam failure rates for honors-level math courses were lower but still significant: 32 percent for geometry and 28 percent for Algebra 2. “It’s not the exams,” one teacher wrote.
You might be interested: Often asked: What Is A Reflection In Geometry?
## Is learning geometry easy?
Geometry is the study of shapes and angles and can be challenging for many students. Many of the concepts are totally new and this can lead to anxiety about the subject. There are a lot of postulates/theorems, definitions, and symbols to learn before geometry begins to make sense.
## Is Geometry for 11th grade?
During their junior year, most students take Algebra II, while others may take Geometry or even Pre-Calculus. Whichever math course your junior high schooler takes, a good 11th grade math curriculum should provide comprehensive knowledge of the core math skills needed for higher education.
## Is Algebra 1 or Geometry harder?
Is geometry easier than algebra? Geometry is easier than algebra. Algebra is more focused on equations while the things covered in Geometry really just have to do with finding the length of shapes and the measure of angles.
## What are the 7 hardest math problems?
The 7 Hardest Math Problems In The World (Unsolved)
1. The Collatz Conjecture.
2. Goldbach’s Conjecture.
3. Twin Prime Conjecture.
4. Riemann Hypothesis.
5. Kissing Number Problem.
6. Unknotting Problem.
7. The Large Cardinal Project.
## Why do I find geometry easier than algebra?
Students who feel as though they prefer to work “visually” with shapes instead of variables in algebraic statements will likely find geometry to be easier than other, more abstract concepts.
## What is the hardest math problem?
But those itching for their Good Will Hunting moment, the Guinness Book of Records puts Goldbach’s Conjecture as the current longest-standing maths problem, which has been around for 257 years. It states that every even number is the sum of two prime numbers: for example, 53 + 47 = 100.
You might be interested: FAQ: What Is The Top Of A Pyramid Called In Geometry?
## Why do I not understand geometry?
For many students, their lack of geometry understanding is due in part from a lack of opportunities to experience spatial curricula. Many textbooks and many district pacing guides emphasize numeracy, arithmetic, and algebraic reasoning. First, there are five sequential levels of geometric thinking.
## Is there dyslexia for math?
Dyscalculia is a condition that makes it hard to do math and tasks that involve math. It’s not as well known or as understood as dyslexia. But some experts believe it’s just as common. That means an estimated 5 to 10 percent of people might have dyscalculia.
## How can I improve my geometry?
Here are 6 ways to ace your geometry homework:
1. Use physical manipulative. The most difficult aspect of geometry is being able to visualize the shape in 3d.
2. Avoid missing classes.
3. Join a study group.
4. Do a lot of practice.
5. Learn from prior mistakes.
6. Answer every question on the homework paper.
## What grade do you take geometry?
Most American high schools teach algebra I in ninth grade, geometry in 10th grade and algebra II in 11th grade – something Boaler calls “the geometry sandwich.”
## Can I teach myself geometry?
You can teach yourself geometry if you master the prerequisites, use a syllabus, have the right resources, get a study buddy, use concept and not rote learning, organize your study materials, allocate time to learn new concepts and time to practice, develop deep work habits, and practice better.
## How do you pass geometry?
Practicing these strategies will help you write geometry proofs easily in no time:
1. Make a game plan.
2. Make up numbers for segments and angles.
3. Look for congruent triangles (and keep CPCTC in mind).
4. Try to find isosceles triangles.
5. Look for parallel lines. | 1,062 | 4,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-33 | latest | en | 0.966704 |
https://quickhomeworksolutions.com/product/a-major-concern-that-arises-with-multiple-regression/ | 1,566,567,166,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318421.65/warc/CC-MAIN-20190823130046-20190823152046-00126.warc.gz | 616,200,844 | 12,122 | +1835 731 5494 Email: instantessays65@gmail.com
A major concern that arises with multiple regression
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A major concern that arises with multiple regression 1) The coefficient of determination is important in explaining variances in estimating equations. For a certain estimating equation, the unexplained variation was given as 22,506. The total variation was given as 51,150. What is the coefficient of determination for the equation? A) 0.34 B) 0.43 C) 0.56 D) 0.75 2) The Perize Corporation used regression analysis to predict the annual cost of indirect materials. The results were as follows: Indirect Materials Cost Explained by Units Produced Constant \$15,685 Standard error of Y estimate \$3,500 r2 0.7832 Number of observations 20 X coefficient(s) 10.25 Standard error of coefficient(s) 2.1876 A) Y = \$15,685 + \$10.25X B) Y = \$3,500 + \$5.15X C) Y = \$19,185 + \$4.48X D) Y = \$12,285 + \$10.25X 3) Corise’s Cola was to manufacture 500 cases of cola next week. The accountant provided the following analysis of total manufacturing costs. Variable Coefficient Standard Error t-Value Constant 115 82.73 1.39 Independent variable 230 105.50 2.18 r2 = 0.82 What is the estimated cost of producing the 500 cases of cola? A) \$115,115 B) \$57,730 C) \$24,380 D) \$9,744 4) Midose’s Stables used two different independent variables (trainer hours and number of horses) in two different equations to evaluate the cost of training horses. The most recent results of the two regressions are as follows: Trainer’s hours: Variable Coefficient Standard Error t-Value Constant \$1,004.65 \$217.93 4.61 Independent Variable \$22.99 \$3.23 7.11 r2 = 0.56 Number of horses: Variable Coefficient Standard Error t-Value Constant \$5,240.95 \$1,180.40 4.44 Independent Variable \$951.48 \$271.85 3.50 r2 = 0.63 What is the estimated total cost for the coming year if 15,000 trainer hours are incurred and the stable has 350 horses to be trained, based upon the best cost driver? A) \$96,327.90 B) \$338,258.95 C) \$345,854.65 D) \$14,277,440.95 5) A major concern that arises with multiple regression is multicollinearity, which exists when ________. A) the dependent variable is not normally distributed B) the standard errors of the coefficients of the individual variables decrease C) the R2 statistic is low D) two or more independent variables are highly correlated with one another Business Management Assignment Help, Business Management Homework help, Business Management Study Help, Business Management Course Help
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There are no reviews yet. | 669 | 2,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-35 | latest | en | 0.798077 |
http://en.wikibooks.org/wiki/Wikijunior:Solar_System/About_gravity,_mass,_and_weight | 1,397,800,230,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00257-ip-10-147-4-33.ec2.internal.warc.gz | 77,224,101 | 12,308 | # Wikijunior:Solar System/About gravity, mass, and weight
Please see About weight and gravity for an alternate version of this module.
Do you know your weight in kilograms or pounds? On Earth, your weight is a number. If you are heavier, then this number is larger. If you go to the moon, or to a space station, is this number the same? Some people say that it is the same, and some people say that it is different. To understand why, you need to know about gravity, mass, and weight.
## Contents
### Mass
Like many words, the word weight can have several different meanings. One meaning of weight is called mass. The word "mass" is used in astronomy and other parts of science.
The mass of an object is simply the amount of material the object is made of. The more material the object is made of, the more mass it has. Things that have a big mass are harder to move and harder to stop than objects with just a little bit of mass. So an empty box (with only air inside) is easier to move than a box filled with books. The box with books has more material, and so more mass, than the empty box.
Our mass is a number for how much stuff there is in our body. That is, if we grow, adding more cells to our body, or retaining more body fat or increasing how big our muscles are, our mass will increase; the number for your mass becomes larger. If we start a diet, and reduce how much fat is kept by our body, the number for our mass becomes smaller. In countries that use the metric system, also called the "Le Système International d'Unités" or SI, the units of mass used for the weight of people is often kilograms (kg).
Suppose that a child has a mass of 40 kg. Now suppose the child goes to the moon or to a space station, but does not change in body composition. Then the number for how much stuff the body has does not change. The child's mass on the moon or at a space station is 40 kg.
Our mass on any planet on the solar system (Jupiter, Venus, Earth, or anywhere else) is the same. But if we step on a scale, the reading will be different on all of these planets, because a scale measures weight, not mass, and weight depends on gravity as well as on mass.
### Gravity
There is another meaning of weight, called "force of gravity". The word "weight" can mean one of two things, "mass" or "force of gravity". But what is "gravity"?
Suppose we jump into the air. We cannot fly, but instead we fall and land on the ground. There is a force which pulls us to the ground. This force is called gravity. The Earth makes gravity, so every time that we jump, we will land on Earth again, because the Earth's gravity pulls us, and we are not strong enough to jump fast enough to escape that gravity.
Which objects make gravity? To make significant gravity, an object must have a very large number for mass. A child may only be 40 kg. The Earth's mass is about 6 million billion billion kg. (That's six followed by twenty-four zeros). The Earth has enough mass to make us fall, as quickly as we do, when we jump.
All things with mass make gravity and attract one another. The more mass an object has, the more it attracts other objects toward it. So while the Earth pulls on a child, the child also pulls on the Earth. The force on the child is the same as the force on the earth. Because the earth is so massive, that force moves it very little. If we could look very closely, though, we would see that when the child jumps, the Earth is pushed away by the force of the child's legs, a very small distance, mostly it's the child who moves. And then, when the child falls, the Earth also "falls" just a little bit toward the child.
Everything with mass makes gravity. When we are on Earth, the moon, the other planets, and the sun are far away, and the force of gravity gets smaller with distance, so we land on Earth again. (Gravity pulls us towards the center of any massive object. Because we are so close to the center of the Earth, compared to the center of other planets and the Sun, we are mostly pulled toward the center of the Earth when we are on the Earth. Gravity would pull us towards the center of the Venus, or any planet, if we were on the Venus or another planet.)
Suppose we went to the moon. Now the Earth is too far away to have much effect on how we would fall. If we jump from the moon, we will land on the moon again. The mass of the moon is about 70 thousand billion billion kg. This is much less than the mass of the Earth, 81 Moons would have as much mass as the Earth.
The force of gravity varies directly with the two masses that we consider, multiplied together. If we are jumping on the moon, our mass would be the same as on the Earth, but the mass of the Earth is 81 times that of the Moon, so the attractive force, if we were at the same distance from the center of the Earth or Moon, would be 1/81 as great for the Moon as for the Earth. Because the Moon is smaller, though, we are closer to the center, and the gravity on the Moon's surface is about one sixth that of the gravity on the surface of the Earth.
To describe how gravity changes with distance is a little more complicated. If we divide the distance by two, we will increase the force by two times two, or four. That's said this way: "The force of gravity varies with the inverse square of the distance." When the distance is large, a little change in distance makes almost no change in the force of gravity, but the change in gravity from moving from a beach near the ocean to the top of a mountain can be measured. It's not enough to notice just from how it feels to jump!
### Force
Gravity exerts a force and that force may be measured in units called "newtons." The force on a mass exerted by gravity will vary with the mass and how strong the gravity is where the mass is located. If we double the mass we will double the force.
However, when we push on an object that is free to move, that object will start to move in the direction we push. How fast it moves depends on three things: how hard we push, the mass of the object, and how long we push.
The more massive the object, the slower will be its response to our pushing. The result is that, although the force of gravity increases with mass, the motion decreases with mass, and these two effects cancel each other out.
So how fast things fall doesn't change with the mass. However, other things may change it, especially friction or air resistance.
A cannonball is pulled more strongly by gravity than a ball-bearing, but it has more mass and takes more effort to start moving. A ball-bearing has less mass, but takes less effort to start moving. Both then take the same amount of time to roll down a ramp, and perhaps you have seen or will see a demonstration of something like this.
Why, then, does something like a feather or piece of paper take so much longer to fall to the ground than a cannonball? This is because the resistance from the air in which they are falling is much greater for the feather or paper than for the cannonball. The resistance is a force opposite in direction to the motion, so it reduces the net force on the object.
(Gravity is always attracting us to the center of the Earth. Even when we are standing on the ground. The ground resists the force of gravity, pushing back up with an equal and opposite force, so we stay put.)
What happens, then, if there is no air? In 1971, astronaut David Scott visited the Moon, where there is no air. He held a feather and hammer, each in one hand, and then dropped them at the same time. They both hit the surface of the moon, also at the same time.
### A table showing how gravity changes what happens elsewhere
If we were able to travel to another world, like the astronauts of the Apollo lunar exploration crews, there are a number of things that we would notice that are different from what you would experience on the Earth. There are also some things that would be just the same.
The following is a table regarding what kinds of experiences, if we weigh 40 kg. on Earth, would have if we visited different planets or moons in the Solar System:
Earth Moon Mercury Venus Mars Phobos
Surface Gravity (compared to Earth = 1000 milli-g) 1000 170 380 900 380 5 (average)
Our weight (mass) 40 kg 40 kg 40 kg 40 kg 40 kg 40 kg
How much we could lift 10 kg 59 kg 26 kg 11 kg 26 kg 2000 kg
How high we could jump 20 cm 120 cm 53 cm 22 cm 53 cm 400 m
Time to fall back to ground (seconds) 0.4 2.4 1.1 0.4 1.1 380
How far we could kick a ball 20 m 120 m 53 m 22m 53 m (into Martian orbit)
In a pressurized chamber like a huge domed city, on the Moon, we would be able to put on wings and flap our arms to fly like birds do here on the Earth. Human powered flight is almost impossible here on the Earth because humans weigh too much here.
Phobos is one of the moons of Mars, and is so tiny that the gravity is very low. For example, if we kicked a ball really hard it could leave Phobos completely and go into orbit as a separate object orbiting Mars. Jumping up would take several minutes before the gravity would pull us back down, so we could jump over a mountain on that moon of Mars if we wanted to.
Of all the objects in the Solar System with a "solid" surface that we could walk on, the Earth has the strongest gravity. Jupiter and Saturn may have stronger gravity, but there is nothing we can say is a "solid" surface to walk on. There may be planets that are larger than the Earth with a solid surface, but they are not found in our Solar System. (However, if there were a floating platform on Jupiter or Saturn, we could walk on it, but it would be difficult, we'd weigh so much.) | 2,234 | 9,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2014-15 | longest | en | 0.944326 |
https://www.gigacalculator.com/converters/convert-cm-to-ft-centimeters-to-feet.php | 1,685,384,447,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00311.warc.gz | 863,901,738 | 14,225 | # Centimeters to Feet Converter
Use this converter to easily convert between Centimeters and Feet (cm to ft).
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## How many Centimeters equal one Foot?
A foot is defined as 0.3048 of a meter, so 30.48 centimeters equal one feet, since 1 cm = 0.01 meters. The symbol for centimeters is "cm", while "ft" is used for feet. Sometimes a single quote does the same job, e.g. 6' tall means 6 feet tall.
## Difference between Centimeters and Feet
Both feet and centimeters are units for measuring length or dimensions of an object. Their magnitude is somewhat different, but there is significant overlap in their application, for example measuring human height, or waist circumference. Both units are used in construction, architecture, landscaping, and so on. However, feet are also used to measure small distances, while centimeters are usually used only for very small distances of up to a couple of meters.
The origin of using the unit of foot is unclear as it goes very far back, likely before writing was invented. It was a part of many local systems of units, including the Greek, Roman, Chinese, French, and English systems of measurement. The length of a foot varied from location to location and sometimes depended on which trade it was applied to, but the length was typically between 250 mm and 335 mm. In medieval Europe the foot was taken as the average of length of the feet of randomly sampled men from a local community. Since 1959 the foot is defined using the metric system. Currently the US is the only country which prefers using the foot in preference to the meter in commerce, engineering and standardization of all kinds.
The centimeter is part of the metric system, adopted in most countries nowadays. The base unit is the meter, which is defined by the distance light travels in vacuum in a fraction of a second. A centimeter is defined as 1/100 of a meter, so 1cm = 0.01m.
## How to convert Centimeters to Feet
Conversion from feet to centimeters requires you to divide the measurement in feet to 30.48. Getting the centimeter equivalent of a foot measurement is much easier using our free online cm to ft converter above.
#### Cm to feet conversion example
Sample task: convert 185 cm to feet. Solution:
Formula:
cm / 30.48 = ft
Calculation:
185 cm / 30.48 = 6.069554 ft
End result:
185 cm is equal to 6.069554 ft
## cm to ft conversion table
cm to ft conversion table
cm ft
1 cm 0.032808 ft
2 cm 0.065617 ft
3 cm 0.098425 ft
4 cm 0.131234 ft
5 cm 0.164042 ft
6 cm 0.196850 ft
7 cm 0.229659 ft
8 cm 0.262467 ft
9 cm 0.295276 ft
10 cm 0.328084 ft
20 cm 0.656168 ft
30 cm 0.984252 ft
40 cm 1.312336 ft
50 cm 1.640420 ft
60 cm 1.968504 ft
70 cm 2.296588 ft
80 cm 2.624672 ft
90 cm 2.952756 ft
100 cm 3.280840 ft
200 cm 6.561680 ft
300 cm 9.842520 ft
400 cm 13.123360 ft
500 cm 16.404199 ft
600 cm 19.685039 ft
700 cm 22.965879 ft
800 cm 26.246719 ft
900 cm 29.527559 ft
1,000 cm 32.808399 ft
#### References
1 NIST Special Publication 330 (2008) - "The International System of Units (SI)", edited by Barry N.Taylor and Ambler Thompson
2 International Organization for Standardization (1993). ISO Standards Handbook: Quantities and units (3rd edition). Geneva: ISO. ISBN 92-67-10185-4.
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# lecture_18 - Calculate the transmission probability T for a...
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x V(x) 0 Calculate the transmission probability T for a particle of mass m and energy E incident from the right on the potential step shown in the figure. The energy E indicated by the dashed line is above the height V 0 of the potential step. Do this calculation by first finding the reflection probability R , then use R+T=1 to calculate the transmission probability T . V 0 E
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7.1 Application of the Schrödinger Equation to the Hydrogen Atom 7.2 Solution of the Schrödinger Equation for Hydrogen 7.3 Quantum Numbers 7.4 Magnetic Effects on Atomic Spectra – The Normal Zeeman Effect 7.5 Intrinsic Spin 7.6 Energy Levels and Electron Probabilities CHAPTER 7 The Hydrogen Atom The Hydrogen Atom The atom of modern physics can be symbolized only through a partial differential equation in an abstract space of many dimensions. All its qualities are inferential; no material properties can be directly attributed to it. An understanding of the atomic world in that primary sensuous fashion…is impossible. - Werner Heisenberg
7.1: Application of the Schrödinger Equation to the Hydrogen Atom The approximation of the potential energy of the electron-proton system is electrostatic: Rewrite the three-dimensional time-independent Schrödinger Equation. For Hydrogen-like atoms (He + or Li ++ ): Replace e 2 with Ze 2 ( Z is the atomic number). Use appropriate reduced mass μ.
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Application of the Schrödinger Equation The potential (central force) V ( r ) depends on the distance r between the proton and electron. Transform to spherical polar coordinates because of the radial symmetry. Insert the Coulomb potential into the transformed Schrödinger equation.
Application of the Schrödinger Equation The wave function Ψ is now a function of ( r , θ, φ ) . Equation is separable.
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# computer notes - Binary tree
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If every non-leaf node in a binary tree has non-empty left and right subtrees, the tree is termed a strictly binary tree. There are a number of operations that can be defined for a binary tree.
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### computer notes - Binary tree
1. 1. ecomputernotes.com Data Structures Lecture No. 12 ___________________________________________________________________ Page 1 of 13 Data Structures Lecture No. 12 Operations on Binary Tree In the last lecture, we talked about the uses of binary tree, which is an abstract data type. We discussed about the functions to find the information inside a node, the parent, the siblings(brothers), the left and right children of a node. In this lecture, we will talk about the algorithms and implementation of those functions. When we discuss about an abstract data type, firstly we focus what it can do for us and don t bother about the how part. The how part or implementation is thought out later. While implementing these abstract data types, the implementation is hidden in a class so it is abstract to the user. The user only needs to be aware of the interface. But there can be situations when the users may like to know about the implementation detail, for example, when a data type is performing quite slower than promised. For now, we start our discussion from the methods of tree data type. Consider a tree has been formed already, following methods will be used to perform different operations on a node of this tree: Operation Description left(p) Returns a pointer to the left sub-tree right(p) Returns a pointer to the right sub-tree parent(p) Returns the father node of p brother(p) Returns the brother node of p info(p) Returns the contents of node p These methods have already been discussed at the end of the previous lecture, however, few more methods are required to construct a binary tree: Operation Description setLeft(p, x) Creates the left child node of p and set the value x into it. setRight(p, x) Creates the right child node of p, the child node contains the info x. All these methods are required to build and to retrieve values from a tree. | 552 | 2,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2016-40 | latest | en | 0.852364 |
https://www.techwhiff.com/issue/select-all-numbers-that-are-in-the-domain--114444 | 1,675,457,661,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500074.73/warc/CC-MAIN-20230203185547-20230203215547-00082.warc.gz | 1,036,269,979 | 11,871 | # Select all numbers that are in the domain.
###### Question:
Select all numbers that are in the domain.
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### Which is a characteristic of a theme? A. It can be expressed in 1 - 2 words. B. It is general and broad. C. It has to do with the story’s setting. D. It expresses a universal idea about life that the reader can relate to.
Which is a characteristic of a theme? A. It can be expressed in 1 - 2 words. B. It is general and broad. C. It has to do with the story’s setting. D. It expresses a universal idea about life that the reader can relate to.... | 1,074 | 4,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-06 | latest | en | 0.939867 |
https://discuss.codechef.com/t/recndnum-editorial/63990/15 | 1,702,336,603,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00027.warc.gz | 242,839,936 | 8,708 | # RECNDNUM - Editorial
How ?
Consider case of n = 1 Till first occurance we have n^2 time elapsed (golden in image). Then onwards look at the points to the left and points to the right.
The left points (red) form mountains of constant height and each containing 2n points.
The right mountains (green) form a sequence 2, 4, 6, 8. Now all that remains is to find out how many of these left and right mountains are required to reach our k.
If you look at editorial expression, the first term is time to reach k = 1. Second term is contribution of left mountains, and third term is contribution of right mountains.
4 Likes
if k is even
(n+(k/2))(n+(k/2))-(k/2)
else
(n+(k/2))
(n+(k/2))+(k/2)
1 Like
See all sequences starting from 0 and ending to 1 as a block. We can get the index of 0 by the formula i*(i+1), where i is the index of the i’th occurence of 0 . Upto the i’th block, the total count of n is 1+(i-n)*2. At each stage, see if this is less than or greater than k and recurse.
General term of sequence where difference of numbers of sequence differ by 2 is `pow(n,2)+ c*n +k` where c and k can be calculated by plugging in some values of n (n denotes the term number of sequence).
Do check out my unofficial editorial Chef and Walk editorial (UNOFFICIAL)
1 Like
Can anyone help me with this plz…which test case will not work with the following code??
ll n,k,x,ans;
cin>>n>>k;
if(n==0)
cout<<((k)*(k-1))%MOD<<endl;
else{
x = (n+k/2)%MOD;
ans = (x * x)%MOD;
if(k&1){
ans = (ans+abs(n-x))%MOD;
}
else{
ans -= abs(n-x);
}
cout<<ans<<endl;
It’s because of overflow
Try this testcase `1000000000 1000000000`
Thanks. But I’m still getting wrong submission with this changed code.
`````` ll n,k,x;
unsigned ll ans;
cin>>n>>k;
if(n==0){
cout<<((k)*(k-1))%MOD<<endl;
}
else{
x = (n+k/2)%MOD;
ans = (x*x)%MOD;
if(k&1){
ans = (ans+abs(n-x))%MOD;
}
else{
ans = (ans-abs(n-x))%MOD;
}
cout<<ans<<endl;
}
``````
Can u tell what I’m doing wrong.
I am not able to find what’s wrong with my code. Here’s my code
https://www.codechef.com/viewsolution/32373259
#define lli long long int
#define mod 1000000007
#include
using namespace std;
#include<bits/stdc++.h>
{
char c;
lli n = 0;
while ((c = getchar_unlocked ()) < 48);
n += (c - ‘0’);
while ((c = getchar_unlocked ()) >= 48)
n = n * 10 + (c - ‘0’);
return n;
}
int main(void)
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
``````lli T=read();
while(T--)
{
// t=n*(n+1) this formula is for total time of nth round (returned back to zero)
lli t=((n%mod)*((n+1)%mod))%mod;
k--;
if(k>0)
{
/*
t=(n+(k/2))*(n+(k/2)+1) same as above but putting k/2 extra rounds with n
i.e. putting n=n+(k/2)
*/
t=(( ((n%mod) + ((k/2)%mod) )%mod) * ((((n%mod) + ((k/2)%mod) +1))%mod));
if(k%2==0)
t=((t%mod)-(n%mod))%mod;//t=t-n
else
t=((t%mod)+(n%mod))%mod;//t=t+n
}
else
t=( (t%mod)-(n%mod))%mod;//t=t-n
cout<<t%mod<<endl;
}
return 0;
``````
}
instead of using long long for n, k. Can we use ‘int’ for them? And only use a long long variable to store the ans. Is this possible or will it overflow then?
Here is my Code for RECNDNUM. It passes all the sample test cases, then where it is going wrong…
https://www.codechef.com/viewsolution/32367147
I tried many random test case generators and compared my code with the Setter’s solution. Outputs are identical in all cases. I don’t know what’s wrong. Any one try and find a test case?
https://www.codechef.com/viewsolution/32381187
Desperate af T_T
heyy geek, please have look on my code.
I used the concept of sum of AP and the first time we find n is n*n
#include
#include
#include <bits/stdc++.h>
#include
using namespace std;
#define ll long long int
int mod=1e9+7;
ll fxp(ll a,ll b,ll m)
{
if(b==2)
return (aa)%m;
if(b%2==0)
return fxp(a
a,b/2,m);
return fxp(a,b-1,m)*a;
}
void swap(ll &a,ll &b){ ll t=a; a=b; b=t;}
int main() {
int t;
cin>>t;
while(t–)
{
ll n,k;
cin>>n>>k;
int res=0;
res=nn;
k=k-1;
if(n==0 && k>0)
{
if(k%2==0)
res+=(k/2)
(4+(k-1)2);
else
res+=((k
(4+(k-1)*2))/2);
`````` }
else
if(k%2==0 && k!=0)
{
res+=(n*2)*k/2;
ll x=k/2;
res+=(x*(4+(x-1)*2))/2;
}
else if(k%2!=0 && k!=0)
{
res+=(n*2)*(k/2)+n*2;
ll x=k/2;
res+=(x/2)*(4+(x-1)*2);
}
cout<<(res%mod)<<endl;
}
return 0;
``````
}
@rishup_nitdgp why this solution (in which I used long long) is giving wrong answer but in setter’s solution which used `int ans = n * n` which is not overflowing whereas in my solution even though I used long long and took modulo at every step then also it resulted in WA.
How come??
My same logic worked in pypy3 so my logic was correct and it was an overflow error then how come when n could be as large as 10^9 and your solution is working?
The ans may be negative when you subtract k/2.
Can anyone tell me which test case will not be executed with this code : -
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const long long MOD = (long long)(1e9 + 7);
ll solve(ll t)
{
ll n,k,start;cin>>n>>k;
if(n==0)
start=1;
else if(n>0)start=n;
if(k==1)cout<<(n*n)%MOD;
else
{k–;
while(k>0)
{start++;
k-=2;
start%=MOD;
if(k<=0)break;
}
if(k<0)
cout<<(((start-1)*start)+n)%MOD;
else
cout<<(((start-1)*start)+start+(start-n))%MOD;
}
``````if(t>0)cout<<"\n";
``````
}
int32_t main()
{
#ifdef ONPC
freopen(“input.txt”, “r”, stdin);
#endif
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
ll t;cin>>t;while(t–)
{
solve(t);
}
``````return 0;
``````
}
But it is working in pypy3.
The point is how the setter’s solution working even though it will definitely overflow for large values of n when ans = n^2. | 1,839 | 5,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-50 | latest | en | 0.833772 |
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### Pseudocode Challenge
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This resource, suitable for upper secondary school students, looks in detail at the night sky, examining objects and orbital motion within the solar system. Constellations and the stories...
### Compose Tune
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This...
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Teachers and session leaders should access the ...
### Starting from Scratch
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### Flowcharts and Pseudocode
An introductory lesson, linking ideas from flowcharts to the use of pseudocode. The presentation may need some editing to remove school specific information. It contains a link to the "Friendship Algorithm" sequence from Big Bang Theory and a link to the description section for Algorithms and Pseudocode from BBC...
### Introduction to Algorithms
This resource details a real-world algorithm which students are unlikely to have encountered previously. The Luhn Algorithm is one method for validating that the long number on a credit/debit card is a valid number (it doesn't check if the card is actually a credit/debit one, just that the number conforms to the...
### Introduction to Pseudocode
This activity features detailed instructions for a lesson to introduce the concept of pseudocode, using fairly simple but accessible graphics. The students are required to devise instructions in order to move a cartoon character on a grid including being able to interact with its' environment by picking up bananas... | 615 | 3,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-51 | latest | en | 0.90401 |
https://gowanusballroom.com/what-is-the-drag-coefficient-of-an-airfoil/ | 1,701,274,682,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100112.41/warc/CC-MAIN-20231129141108-20231129171108-00818.warc.gz | 340,791,061 | 11,428 | ## What is the drag coefficient of an airfoil?
The drag coefficient is a number which aerodynamicists use to model all of the complex dependencies of drag on shape, inclination, and some flow conditions. The drag coefficient Cd is equal to the drag D divided by the quantity: density r times reference area A times one half of the velocity V squared.
What NACA 0015?
The NACA 0015 airfoil is symmetrical, the 00 indicating that it has no camber. The 15 indicates that the airfoil has a 15% thickness to chord length ratio: it is 15% as thick as it is long.
### Which aircraft uses NACA 4412?
The NACA 4412 has been used in sports plane. The example includes AAI-AA2 mamba aircraft, aeronca series aircraft like aeronca 65-tac defender, aeronca 11ac chief etc while the S1223 is an airfoil used in heavy lift cargo planes [2].
What is a good drag coefficient?
Typical drag coefficients The average modern automobile achieves a drag coefficient of between 0.25 and 0.3. Sport utility vehicles (SUVs), with their typically boxy shapes, typically achieve a Cd=0.35–0.45. The drag coefficient of a vehicle is affected by the shape of body of the vehicle.
## Is a higher drag coefficient better?
A low coefficient is conducive to high top speed and low fuel consumption, while a higher drag coefficient is generally found in cars searching for high cornering speeds influenced by downforce. It shows that a car’s coefficient of drag can be found by analysing the drag force acting on the car at a given speed.
How do you read NACA 5 Series?
NACA Five-Digit Series: The first digit, when multiplied by 3/2, yields the design lift coefficient (cl) in tenths. The next two digits, when divided by 2, give the position of the maximum camber (p) in tenths of chord. The final two digits again indicate the maximum thickness (t) in percentage of chord.
### How does the NACA 0015 airfoil perform for lift and drag analysis?
The NACA 0015 airfoil was analyzed for the lift, drag and moment coefficients as planned. The measured values determined from lab data agree correctly with the theoretical values for the lift, drag and quarter chord moment. A stabilizing or restoring moment was observed after the stall occurred.
What is the value of lift and drag coefficient?
LIFT AND DRAG COEFFICIENTS Lift coefficient is a dimensionless quantity which is defined as = 2 (1) 2
## What are the factors that affect the accuracy of NACA results?
up to 1. 60. Reduction of Data The test data have been reduced to standard NACA coefficient form. Factors which could affect the accuracy of these results, together with the corrections applied, are discussed in the following paragraphs. Tunnel-wall interference.
What is the NACA 0015?
The NACA 0015 is a symmetrical airfoil with a 15% thickness to chord ratio. Symmetric airfoils are used in many applications including aircraft vertical stabilizers, submarine fins, rotary and some fixed wings. A 2D wing section is analyzed at low speeds for lift, drag and moment characteristics. | 694 | 3,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-50 | latest | en | 0.921798 |
https://learningplus.ph/products/life-of-fred-elementary-mathematics-dogs | 1,670,214,036,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711003.56/warc/CC-MAIN-20221205032447-20221205062447-00444.warc.gz | 403,344,992 | 21,492 | ## Life of Fred - Elementary Mathematics: Dogs
• ₱750.00
You are ready to start Life of Fred Elementary when...
1. You can talk in sentences.
2. You are willing to listen to a 5 page story.
3. You can hold a pencil.
4. You can print the digits 0 to 10
5. You know your basic addition and subtraction facts up to 10
Dr. Schmidt recommends that all students up to 4th grade start with Apples and work their way through the entire elementary series.
This book covers beginning mathematics including . . .
• One Million
• the Game of Doubles
• Fortnight
• Digits in a Number
• Two Dimes = 20 Cents
• Doubling 1 ➆2 ➆4 ➆up to 1,267,650,600,228,229,401,496,703,205,376 (which is 100 doubles)
• 7 - z = 4
• Functions
• the Function Party Game
• One Billion
• Making Choices in Life
• Finding Patterns
• Consecutive Even Numbers
• Numbers that Add to 15
• Guessing Functions
• Right Angles
• Rectangles
• Constant Functions
• Numbers that Add to 17
• Bar Graphs
• Adding two- and three-digit numbers
• Carrying the One in Addition
• and much more . . .
Unlike all other math programs, this one also teaches about:
Our Place in the Dance of Life
New Words in English
Anachronism
Middle English
Old English
Making Choices in Life
Daniel Defoe's Robinson Crusoe
Morse Code from A–Z
What Can Cause Unclear Thinking
Rhyme Schemes
Tennyson's In Memoriam
ATM Cards
Sheet Music for Borrowed Books
Beautiful Handwriting
Macronutrients
The Eight Planets
Seven Wonders of the Ancient World
Different Ways to Pick Out a Book to Read
Different Jobs You Might Choose
Idioms
Spendthrifts
Female and Male
Apartment Leases
Isotopes of Hydrogen
The Chemical Elements
Dog Games
The answers are included in the textbook.
Life of Fred Dogs is a hardcover textbook containing 128 pages. This book is not consumable. All answers are written on separate paper or in a notebook. It is designed to be read with the student.
Life of Fred Dogs contains 19 lessons and is designed to take approximately one month to complete.
(description c/o the Life of Fred website)
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Monday 18th of December
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9th grade algebra study guide
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Registered: 02.12.2005
From:
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https://math.answers.com/math-and-arithmetic/What_is_the_answer_to_this_equation_5t-26_equals_equals_18t | 1,679,538,596,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00636.warc.gz | 431,494,699 | 49,741 | 0
# What is the answer to this equation 5t-26 equals equals 18t?
Wiki User
2011-04-23 14:51:10
5t-26 = 18t
-26 = 18t -5t
-26 = 13t
t = -2
Wiki User
2011-04-23 14:51:10
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2240 Reviews | 106 | 234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-14 | longest | en | 0.825265 |
https://www.physicsforums.com/threads/where-will-the-center-of-gravity-of-the-wooden-plank-be.432882/ | 1,726,679,522,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00407.warc.gz | 842,420,032 | 19,674 | # Where Will the Center of Gravity of the Wooden Plank Be?
• zorro
In summary: The Attempt at a SolutionIn summary, the centre of mass of the plank is the mid point of the plank and the torque on both sides is equal.
zorro
## Homework Statement
A thin wooden plank 1m long is kept on the protruding stone with some of its part immersed in water.
Will the centre of gravity of the wooden plank be inside the water or outside?
## The Attempt at a Solution
For the wooden plank to be in equilibrium, moments of buoyancy force and weight about the point of contact of stone should be equal.
Buoyancy force acts at the midpoint of the immersed part.
But what will be the position of the centre of mass (gravity) of the plank?
#### Attachments
• rod.jpg
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How long is the rod?
1 metre
The position of the centre of mass (gravity) of the plank is the mid point of the plank.
rl.bhat said:
The position of the centre of mass (gravity) of the plank is the mid point of the plank.
then does it mean the torque on both side is equal?
rl.bhat said:
The position of the centre of mass (gravity) of the plank is the mid point of the plank.
What I asked was whether it will should lie inside the water or outside to get balanced?
Assume that x-axis is oriented along the thin plank and area of cross section of the plank 1 cm^2.
Density of wood = 0.7 g/cm^3 and that of water = 1 g/cm^3
Let x cm be the portion of the plank in the water.
Length of the plank is 100 cm. So the mass of the plank = 70 g.
Mass of the displaced water = x g. Then buoyancy force is proportional to x.
If R is the reaction of the stone, R = (70 - x)
Take the moment about the end of the plank out side the water.
R*10 + x(100 - x/2) = 70*50
Solve the quadratic. If x is more than 50, C.G. is inside the water. Otherwise it is outside the water.
How did you get R = 70-x ?
I have assumed that the component of the weight of the plank perpendicular to x-axis is
L*d*g*cosθ. Length of the plank is 100 cm, density 0.7 g/cm^3
Component of the weight of the displaced liquid = x*1*g*cosθ
For equilibrium, total downward with respect x-axis must be equal to the total upward force. Weight is in the downward direction and buoyancy and the reaction are in the upward direction.
Hence R*g*cosθ = 70*g*cosθ - x*g*cosθ
Last edited:
Thanks!
## 1. What is buoyancy?
Buoyancy is the upward force exerted by a fluid on an object immersed in it. It is caused by the difference in pressure between the top and bottom of the object.
## 2. How does an object's shape affect its buoyancy?
An object's shape affects its buoyancy because it determines the amount of fluid it displaces. Objects that displace more fluid will experience a greater buoyant force.
## 3. What is the center of gravity?
The center of gravity is the point at which an object's weight is evenly distributed in all directions. It is the point where an object will balance perfectly if suspended.
## 4. How does an object's center of gravity affect its stability?
An object's center of gravity plays a crucial role in its stability. If the center of gravity is located above the object's base, it will be stable, but if it is outside the base, the object will be prone to tipping over.
## 5. How are buoyancy and center of gravity related?
Buoyancy and center of gravity are closely related because the position of an object's center of gravity affects its buoyancy. If the center of gravity is above the object's center of buoyancy, it will be stable and remain upright. However, if the center of gravity is below the center of buoyancy, the object will be top-heavy and may tip over.
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2K | 991 | 3,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.903376 |
https://math.stackexchange.com/questions/2921853/coupon-collecting-problem-4-coupons-with-p-1-p-2-frac18-and-p-3 | 1,601,078,139,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228998.45/warc/CC-MAIN-20200925213517-20200926003517-00035.warc.gz | 493,487,202 | 37,828 | # Coupon Collecting Problem: $4$ coupons with $p_1 = p_2 = \frac{1}{8}$ and $p_3 = p_4 = \frac{3}{8}$
This is from Ross.
I know how to solve everything but (d).
The book answer is $\frac{123}{35}$.
There are $4$ different types of coupons, the first $2$ of which compose one group and the second $2$ another group. Each new coupon obtained is type $i$ with probability $p_i$, where $p_1 = p_2 = 1/8$ and $p_3 = p_4 = 3/8$. Find the expected number of coupons that one must obtain to have at least one of
(a) all 4 types; (b) all the types of the first group; (c) all the types of the second group;
(d) all the types of either group.
My attempt
Let $X =$ number of coupons needed to collect all the types of either group.
Let's number the coupons $1$ through $4$ where $1$ and $2$ are part of the first group and $3$ and $4$ are part of the second. Let's say that the notation $4132$ represents the order of newest types seen. For example if you observe $44111332$ then the order of newest types seen was $4132$.
To get the $E[X]$ I'll condition on every arrangements where the game stops once you have all the types of either group.
$$E[X] = E[X \mid 12]P(12) + E[X \mid 132]P(132) + E[X \mid 134]P(134) + E[X \mid 142]P(142) + E[X \mid 143]P(143) + E[X \mid 21]P(21) + E[X \mid 231]P(231) + E[X \mid 234]P(234) + E[X \mid 241]P(241) + E[X \mid 243]P(243) + E[X \mid 312]P(312) + E[X \mid 314]P(314) + E[X \mid 321]P(321) + E[X \mid 324]P(324) + E[X \mid 34]P(34) + E[X \mid 412]P(412) + E[X \mid 413]P(413) + E[X \mid 421]P(421) + E[X \mid 423]P(423) + E[X \mid 43]P(43)$$
For example to get the first term, $E[X \mid 12] = 1 + \frac{2}{1}$ where the logic is given that you know the coupon types will appear as $12$, the number of coupons needed to get $1$ must be one and then to get $2$ it's the expected value of a geometric RV with $p=\frac{1/8}{1/8+1/8} = \frac{1}{2}$.
To get $P(12)$ we use the multiplication rule. $P(12) = P(1) P(1 \mid 2) = \frac{1}{8} \frac{1}{7} = \frac{1}{56}$. So $E[X \mid 12]P(12) = (1 + \frac{2}{1}) \frac{1}{56}$.
To get the second term, $E[X \mid 132] = 1 + \frac{4}{3} + \frac{4}{1}$ where you know the number of coupons needed to collect $1$ and $2$ is one and to get the middle it's a geometric RV with $p = \frac{3/8}{3/8+1/8} = 3/4$.
To get $P(132) = P(1) P(3 \mid 1) P(2 \mid 13) = \frac{1}{8} \frac{3}{7} \frac{1}{4}$.
Next put it in a python script and get the answer.
Unfortunately I get $3.4 \neq \frac{123}{35} = 3.51428$.
My questions
Can you point out where I'm going wrong? Also, what is the more elegant approach? Thanks.
• FYI - the two different approaches below by Ross M and joriki are great and work. The attempt written in the original post by me is wrong and also the least elegant but it can be fixed. $E[X \mid 12] = 1 + \frac{1}{1-\frac{1}{8}} = 15/7$ and not $1+\frac{2}{1}$. (Thanks to joriki for pointing out how to fix it in question 2923059) – HJ_beginner Sep 21 '18 at 5:06
I would say there are five states, $a,b,c,d,e$, where $a$ has no coupons, $b$ has one coupon from group $1$ and none from group $2$, $c$ has one coupon from group $2$ and none from group $1$, $d$ has one coupon from each group, and $e$ has two coupons from one group and is final.
If you are in state $d$ you finish with probability $\frac 12$ so the expected number of draws from $d$ is $2$.
If you are in state $c$ you finish with probability $\frac 38$, go to $d$ with probability $\frac 14$ and stay in $c$ with probability $\frac 38$. The expected number of draws from $c$ is $$E(c)=1\cdot \frac38 + (1+E(d))\cdot \frac14 +(1+E(c))\cdot \frac 38\\ \frac 58E(c)=\frac 32\\E(c)=\frac {12}5$$ If you are in state $b$ you finish with probability $\frac 18$, go to $d$ with probability $\frac 34$ and stay in $b$ with probability $\frac 18$. The expected number of draws from $b$ is $$E(b)=1\cdot \frac 18+(1+E(d))\cdot \frac 34+(1+E(b))\frac 18\\ \frac 78E(b)=\frac 52\\E(b)=\frac {20}7$$ Finally from $a$ you go to $b$ with probability $\frac 14$ and to $c$ with probability $\frac 34$ so $$E(a)=(1+E(b))\frac 14+(1+E(c))\frac 34\\=\frac {27}{28}+\frac {51}{20}\\=\frac {123}{35}\approx 3.514$$
• No, I am not the same Ross that is the source of the problem. – Ross Millikan Sep 18 '18 at 18:30
• Thanks for your solution... very clever... it makes sense though I will need to meditate on it. Small point, but I think the final equality should be an approximation... that is $\frac{27}{28}+\frac{51}{20} \approx \frac{25}{7}$ but whatever you have the right answer. By any chance do you have any thoughts on what was wrong with my approach? Instead of thinking about it in general states like you did I just conditioned on every state. – HJ_beginner Sep 18 '18 at 18:47
• Actually, $\frac {27}{28}+\frac {51}{20}={123\over 35}$ the same answer the other Ross gets. – saulspatz Sep 18 '18 at 18:56
• @saulspatz: thanks. I misrecognized the decimal when I calculated the sum – Ross Millikan Sep 18 '18 at 19:14
• When you compute E(132) the waiting time for the $2$ is $5$ because you are guaranteed to draw a $1,2,$ or $3$ and the $2$ accounts for $\frac 15$ of the probability. – Ross Millikan Sep 18 '18 at 19:32
Let $N_1$, $N_2$, $N_\land$ and $N_\lor$ denote the numbers of coupons you need to draw to get all types of the first group, of the second group, of both groups and of either group, respectively. Then
\begin{eqnarray*} \mathsf E[N_\lor] &=& \sum_{n=0}^\infty\mathsf P(N_\lor\gt n) \\ &=& \sum_{n=0}^\infty\mathsf P(N_1\gt n\land N_2\gt n) \\ &=& \sum_{n=0}^\infty\left(\mathsf P(N_1\gt n)+\mathsf P(N_2\gt n)-\mathsf P(N_1\gt n\lor N_2\gt n)\right) \\ &=& \sum_{n=0}^\infty\left(\mathsf P(N_1\gt n)+\mathsf P(N_2\gt n)-\mathsf P(N_\land\gt n)\right) \\ &=& \mathsf E[N_1]+\mathsf E[N_2]-\mathsf E[N_\land]\;. \end{eqnarray*}
These are the results of parts (a) through (c) that you've already solved; the result is
$$\mathsf E[N_\lor]=12+4-\frac{437}{35}=\frac{123}{35}\;.$$
• Hmmm very nice... I figured when doing part (d) that the results from part (b) and (c) would come into play. It looks like you used inclusion exclusion rule and the one property for expected value (your first equality) that I would not be clever enough to think of using for this problem. Question... should the second row be $N_1 > n \cup N_2 > n$? Where it's $\cup$ vs $\cap$? I could be totally off... – HJ_beginner Sep 18 '18 at 21:00
• @HJ_beginner: No, I think it's the right way around. We haven't completed either group if we haven't completed the first group and we haven't completed the second group. As regards thinking of this property of the expected value: Keep that in mind; it often comes in useful in coupon collection problems. I went through my answers to coupon collection problems and found that $7$ out of $32$ (including this one) use this property: – joriki Sep 18 '18 at 21:16
• math.stackexchange.com/a/1842984, math.stackexchange.com/a/1824331, math.stackexchange.com/a/1454749, math.stackexchange.com/a/1836308, math.stackexchange.com/a/1786978, math.stackexchange.com/a/1794043 (where the last one confirms the connection you draw to inclusion-exclusion) – joriki Sep 18 '18 at 21:17
• As always thank you for your help. I will check those out! I believe your answer is in line with what Sheldon Ross (the author) was thinking. But I already gave other Ross the check mark and his answer was great too. I would give multiple checks if I could. #greencheck – HJ_beginner Sep 18 '18 at 22:10
• @HJ_beginner: You might also want to check out this answer, where I reapplied the two approaches Ross and I took here. Whereas I agree that here my approach is probably what Sheldon Ross was aiming for, in that other case Ross's approach requires quite a bit less computation than mine. – joriki Sep 20 '18 at 17:01 | 2,600 | 7,798 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-40 | latest | en | 0.814277 |
http://www.jiskha.com/display.cgi?id=1402231488 | 1,462,007,429,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111809.10/warc/CC-MAIN-20160428161511-00137-ip-10-239-7-51.ec2.internal.warc.gz | 602,844,796 | 3,540 | Saturday
April 30, 2016
# Homework Help: physics
Posted by Vidhya on Sunday, June 8, 2014 at 8:44am.
What is the pressure due to a man weighing 80 kg standing on his feet? given area of his feet is 160sq.cm take g=10m/s^2
## Answer This Question
First Name: School Subject: Answer:
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1. I've read different views on this, but should you take a break during the NCLEX?
•
Joined: Aug '10; Posts: 3; Likes: 1
from US
3. If you need one, you need one. What if you get an urgent urge to pee? You would take a break for that wouldn't you?
4. if you still have lots of time and you think it's enough for 265 questions (nobody knows when will the computer stop right?) then take a break.
i tried w/o any breaks and staring at the computer screen hurts my eyes that i'm having a hard time reading and kept wiping my tears. got a bid bored and anxious too to move away from the crampy chair cost some mistakes for me...
5. Well, if you didn't take any breaks at all and you got all 265 questions you would have about 1.35 minutes per question. Let's say you took a break every 50 questions, that would be about 5 breaks. Let's say each break is 5 minutes, except for 1, which is 10 minutes. That would be a total of 30 minutes (4 breaks x 5 mins = 20 mins + 10 mins = 30 minutes). So that would give you about 1.24 minutes per question. So, I guess it depends on how many breaks you need vs. how many seconds/minutes you need per question to feel comfortable.
Maybe you should take some mock nclex tests at home and time yourself... see how many questions you can do before you get tired/bored/or whatever else. Take about a 5 minute break when you need it and see how long it takes you... that might give you some insight as to how many questions you can do before you want a break and also tell you how long it would take you to do 265 questions to see if you are within the 6 hour time limit. Then monitor and adjust!
Last edit by irish_rainbow on Aug 29, '10 : Reason: Error in post
6. Definitely take a break if you need it. If you feel your heart beating hard and can hear your heart beating through the noise-reduction head phones then yes. Take a breather, knock out the bathroom while you're at it, refocus and finish up. | 490 | 1,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-22 | latest | en | 0.964284 |
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kokoory August 17, 2011 08:20
Accelerating pre and checking the mfr at exit how to hook udf to Pressure inlet(BC)
hello, guys.
I have a problem to apply udf to my model.
the problem is analysizing very small nozzle.
However, I have no idea to code and apply udf.
first, I used a pressure inlet to vary a total pressure from 0 to 10Mpa using udf.
It's good to check and analysis the nozzle. But, in addition, I have to apply checking a mass flow rate at nozzle exit. after checking the mass flow rate, I should apply a volumetric heat addition as per mass flow rate( mass flow rate * mdot Q).
here is the question.
1. I hooked the below code, however, I think it is not working. I wanna do change the pressure from 2Mpa to 10Mpa and check the mass flow rate at the exit simultaneously.
plz help me..
#include "udf.h"
face_t f;
cell_t c;
DEFINE_EXECUTE_AT_END(mass)
{
Domain *d;
real flow=0;
real p=0;
d = Get_Domain(1);
begin_f_loop(f,t)
{
flow+=F_FLUX(f,t);
}
end_f_loop(f,t)
p = C_P(c,t);
printf("MASS Flow Rate: %g\n",flow);
}
DEFINE_PROFILE(pre, t, i)
{
real pressure_mag;
real cu_tim;
//face_t f;
cu_tim = CURRENT_TIME;
if((cu_tim>=0)&&(cu_tim<10))
{
pressure_mag=202650+101325*cu_tim;
}
else
{
pressure_mag=1013250;
}
begin_f_loop(f,t)
{
F_PROFILE(f,t,i) = pressure_mag;
}
end_f_loop(f,t)
}
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# 6 Tablespoons of Butter Equals How Many Cups
Published
on
I’ve always been a stickler for precise measurements in the kitchen.
So, when a recipe called for 6 tablespoons of butter, I wondered how many cups that would be.
After some research and calculations, I discovered the conversion ratio and the answer.
In this article, I’ll share with you the exact measurement and some tips for accurately measuring butter in cups.
So, let’s dive in and simplify your cooking process!
## Key Takeaways
• 6 tablespoons of butter is equal to 0.375 cups
• Knowing the conversion ratios between tablespoons and cups ensures precise measurements when using butter in recipes
• Using a kitchen scale for accurate measurement of ingredients is recommended
• Experimenting with alternative ingredients like applesauce, mashed bananas, avocado, or coconut oil can be done when substituting butter
## Understanding the Conversion Ratio
I understand how the conversion ratio works – 1 cup of butter is equal to 16 tablespoons. When measuring butter in grams, it’s important to know how to convert it from sticks to cups accurately.
One stick of butter weighs approximately 113 grams or 4 ounces. To convert this to cups, you need to know that there are 8 tablespoons in a stick of butter. So, if you’ve 1 stick of butter, you’ve 8 tablespoons, which is equivalent to half a cup.
If you’ve 2 sticks of butter, you’ve 16 tablespoons, which is equal to 1 cup. It’s essential to know these conversions to ensure precise measurements when cooking or baking with butter.
## Calculating Butter in Cups
To convert tablespoons of butter to cups, I’d simply divide the amount by 16. When it comes to converting butter measurements in metric units, it’s important to remember that 1 cup of butter is equal to 16 tablespoons. So, if you’ve 8 tablespoons of butter, you’d divide 8 by 16 to get 0.5 cups.
This conversion is essential when following recipes that call for butter in cups rather than tablespoons. Additionally, if you prefer to use butter substitutes in your recipes, such as margarine or oil, it’s crucial to know the conversion ratio between these substitutes and butter. Keep in mind that different substitutes have different ratios, so it’s always best to consult a reliable conversion chart or follow the instructions provided on the substitute’s packaging.
Understanding these conversions will ensure accurate measurements in your recipes and help you achieve the desired results. Now, let’s move on to some tips for accurate measurement.
## Tips for Accurate Measurement
Using a kitchen scale is a precise way to measure ingredients for accurate recipe results. It ensures that you’re using the exact amount of each ingredient, which is crucial for achieving the desired texture and flavor in your baked goods.
When it comes to measuring butter, many people make common mistakes that can affect the outcome of their recipes. One common mistake is using a measuring cup to measure solid butter, resulting in inaccurate measurements. Instead, it’s best to use a kitchen scale to measure butter in grams or ounces.
Another mistake is using cold butter when the recipe calls for softened butter. This can affect the texture of your baked goods.
If you don’t have butter on hand or prefer a healthier alternative, there are several options you can use in baking. Some alternatives for butter include applesauce, mashed bananas, avocado, or coconut oil. These substitutions may slightly alter the taste and texture of your baked goods, so it’s important to experiment and adjust the recipe as needed.
## Converting Recipes With Butter Measurements
When converting recipes, it’s important to accurately measure the amount of butter needed for the desired outcome. In baking, butter adds flavor, moisture, and richness to our favorite treats. However, there may be times when we need to use butter substitutes for dietary reasons or simply because we ran out of butter. In such cases, it’s crucial to know the proper conversions to ensure our recipes turn out just right.
To measure butter without a scale, you can use the tablespoon method. One stick of butter is equal to 8 tablespoons or 1/2 cup. So, if your recipe calls for 1 cup of butter, you’ll need 2 sticks or 16 tablespoons.
Now that we’ve covered butter measurements, let’s move on to common culinary conversions.
## Common Culinary Conversions
I often find it helpful to have a list of common culinary conversions on hand for quick reference in the kitchen. When it comes to baking recipes that call for butter, it’s important to know how to convert butter measurements accurately.
Sometimes, you may need to substitute butter with a different ingredient, like oil or margarine. To convert tablespoons of butter to cups, remember that 1 cup of butter is equal to 16 tablespoons. So, if a recipe calls for 8 tablespoons of butter, you’d need half a cup.
It’s crucial to get these conversions right to ensure the perfect texture and flavor in your baked goods. Having a list of butter substitutes and knowing how to convert butter measurements will make your time in the kitchen much easier.
### Can I Use Margarine Instead of Butter in Recipes That Call for Cups of Butter?
Sure, you can use margarine instead of butter in recipes that call for cups of butter.
However, keep in mind that margarine and butter have different properties. Butter contains more fat and has a rich flavor, while margarine is made from vegetable oils and has a milder taste.
### How Do I Convert Tablespoons of Butter to Grams?
When it comes to converting tablespoons of butter to grams, I’ve got you covered. Let’s skip the context and dive right into the current question.
Converting tablespoons to grams can be a bit tricky, but fear not! There are handy conversion charts available that provide precise measurements for butter. These charts will make your life so much easier when it comes to baking and cooking. Trust me, they’re a game-changer.
### What Is the Equivalent Measurement for 6 Tablespoons of Butter in Ounces?
To find the equivalent measurement for 6 tablespoons of butter in ounces, you’ll need to convert it. I can help you with that!
Converting tablespoons to ounces is pretty straightforward. One tablespoon is equal to 0.5 ounces.
So, if you’ve 6 tablespoons of butter, it would be equivalent to 3 ounces.
### Are There Any Substitutions for Butter if I Want to Make a Dairy-Free Recipe?
When it comes to making a dairy-free recipe, there are several options you can use as substitutes for butter.
One popular choice is dairy-free butter substitutes, which are made from plant-based oils. These substitutes can be used in equal amounts as butter in most recipes.
Additionally, there are other tips for baking without butter, such as using coconut oil or applesauce as alternatives.
### How Can I Accurately Measure Butter Without Using Tablespoons or Cups?
When it comes to accurately measuring butter without using tablespoons or cups, there are a few alternative methods. One approach is to use the markings on the butter wrapper itself. Each stick of butter usually has measurements marked in tablespoons and cups.
Another option is to use a kitchen scale to weigh the butter. This method allows for precise measurements and eliminates the need for volume-based measurements.
## Conclusion
In conclusion, understanding the conversion ratio between tablespoons and cups is essential for accurate measurement in cooking.
By knowing that 6 tablespoons of butter is equal to 1/3 cup, you can easily adapt recipes and achieve the desired results.
Remember to use proper measuring tools and techniques for precise measurements.
With this knowledge, you can confidently convert recipes and create delicious dishes without any confusion.
# How to Make Butter Sauce for Crabs
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I’ve always loved indulging in a plate of succulent crabs, especially when they’re smothered in a rich and flavorful butter sauce.
If you’re like me and crave that perfect blend of creamy and tangy, then you’ve come to the right place. In this article, I’ll share my tried-and-true method for creating a mouthwatering butter sauce that will elevate your crab feast to new heights.
## Key Takeaways
• Butter, garlic, lemon juice, and parsley are key ingredients for making butter sauce for crabs.
• Melting butter correctly is important for achieving the desired taste and texture of the sauce.
• Infusing flavors into the butter sauce can be done by using fresh herbs, sautéing garlic, and adding citrus zest.
• Steaming crabs is a popular cooking method that retains the natural flavors and textures of the meat, while also preserving moisture and allowing the natural sweetness of the crab to shine through.
## Ingredients for Butter Sauce
First, you’ll need to gather the necessary ingredients for your butter sauce. These include butter, garlic, lemon juice, and parsley.
Butter sauce is a classic accompaniment to crabs. It enhances their natural flavors with its rich and creamy texture.
However, there are different variations of butter sauce that you can try to add a twist to your dish.
One alternative is substituting butter with olive oil for a healthier option. This still provides a smooth and velvety base for your sauce.
Another alternative is using ghee, which is clarified butter and has a nuttier flavor.
Additionally, you can experiment with different herbs and spices to customize the taste of your butter sauce. Try adding dill or thyme for a refreshing twist, or even a dash of cayenne pepper for a spicy kick.
The possibilities are endless when it comes to creating your own unique butter sauce for crabs.
## Preparing the Butter Base
When it comes to preparing the perfect butter sauce, understanding the techniques for melting butter is crucial.
Whether you prefer the stovetop method or the microwave approach, knowing the right way to melt butter can make a significant difference in the final taste and texture of your sauce.
Additionally, exploring different flavoring options for butter, such as herbs, spices, or citrus zest, can elevate the taste of your sauce and add a unique twist to your dish.
### Butter Melting Techniques
To melt the butter for your sauce, you’ll want to start by placing it in a microwave-safe bowl and heating it on high for 30-second intervals, stirring in between, until it is completely melted. This method is quick and convenient, ensuring that your butter is perfectly melted and ready for use in your delicious crab sauce.
Another technique you can try is the clarified butter technique, which involves melting the butter slowly on low heat, allowing the milk solids to separate from the fat. This results in a pure and clear butter, perfect for butter basting your crab. Simply melt the butter in a saucepan over low heat, skimming off the foam and milk solids that rise to the surface.
### Flavoring Options for Butter
For even more variety, you can experiment with different seasonings and herbs to enhance the flavor of your melted butter. Savory seasonings such as garlic powder, onion powder, and Old Bay seasoning can add a delicious kick to your butter sauce. If you prefer a sweeter taste, consider adding a touch of honey or maple syrup. These sweet additions can complement the natural sweetness of the crab meat and create a delightful balance of flavors. Additionally, you can try incorporating fresh herbs like parsley, thyme, or dill to add a burst of freshness to your sauce. The table below provides a list of suggested seasonings and herbs to liven up your butter sauce for crabs.
Savory Seasonings Sweet Additions Fresh Herbs
Garlic powder Honey Parsley
Onion powder Maple syrup Thyme
Old Bay seasoning Dill
## Infusing Flavors Into the Sauce
Adding herbs and garlic is a great way to infuse flavors into the butter sauce for crabs. The infusion techniques enhance the flavors and make the sauce more delicious.
Here are three ways to infuse flavors into the butter sauce:
• Fresh herbs: Chop up some fresh herbs like parsley, thyme, or dill and add them to the melting butter. The heat will release their aromatic oils, infusing the sauce with a fragrant and savory taste.
• Garlic: Crush a few cloves of garlic and sauté them in the butter before adding other ingredients. The garlic will infuse the sauce with a rich and slightly pungent flavor.
• Citrus zest: Grate the zest of a lemon or orange and sprinkle it into the sauce. The citrus oils will add a bright and tangy note to the buttery goodness.
## Cooking the Crabs for the Sauce
When it comes to cooking crabs for my sauce, I have two main options: simmering or boiling. Each method has its own advantages and it ultimately depends on the flavor and texture I’m looking for.
Additionally, I need to consider the timing for cooking the crabs to ensure they are perfectly cooked and ready to be enjoyed with my delicious butter sauce.
### Simmer or Boil Crabs
To simmer or boil the crabs, you’ll need a large pot and enough water to fully submerge them. This step is crucial in cooking the crabs properly and ensuring that they are cooked through.
Both boiling and simmering are common methods used to cook crabs, but which is the best cooking method? Let’s break it down:
• Boiling: This method involves cooking the crabs in rapidly boiling water. It’s a quick and efficient way to cook the crabs, resulting in tender and flavorful meat. However, boiling can sometimes cause the crab shells to crack and the meat to become waterlogged.
• Simmering: Simmering the crabs involves cooking them in gently simmering water. This method is more gentle and can help prevent the shells from cracking. The meat tends to retain its natural flavors and texture.
Ultimately, the best cooking method for crabs depends on personal preference. If you prefer tender meat, boiling may be the way to go. If you want to preserve the natural flavors and texture, simmering is a great option.
### Steaming or Boiling Method
The steaming method requires less water than boiling and can result in a more delicate flavor and texture for the crabs. When steaming crabs, you can use a steamer basket or a large pot with a rack. Fill the pot with about an inch of water and bring it to a boil. Place the crabs in the steamer basket and set it over the boiling water. Cover the pot and steam the crabs for about 15-20 minutes, or until they turn bright red. Steaming preserves the natural sweetness of the crab meat and prevents it from becoming waterlogged. It also helps to retain the vibrant color of the shell. Steamed crabs can be enjoyed on their own or served with melted butter and a variety of seasonings, such as Old Bay, garlic butter, or lemon pepper. This method is a healthier alternative to frying and allows the true flavors of the crab to shine through.
Method Water Usage Flavor/Texture
Steaming Requires less water Delicate flavor and texture
Boiling Requires more water Stronger flavor, softer texture
Frying No water required Crispy texture, added oil flavor
Steaming vs Frying:
Steaming crabs is a popular method for cooking them, as it results in a more delicate flavor and texture compared to frying. Frying involves submerging the crabs in hot oil, which can make them greasy and heavy. Steaming, on the other hand, allows the natural flavors of the crab to shine through without adding any extra oil. It also helps to retain the moisture in the meat, making it tender and succulent.
Seasoning Options:
When steaming crabs, you can enhance their flavor by adding various seasonings. One popular choice is Old Bay seasoning, a classic blend of herbs and spices that pairs perfectly with seafood. You can also try garlic butter, which adds a rich and savory taste to the crabs. For a tangy twist, sprinkle some lemon pepper on the steamed crabs. The options are endless, and you can experiment with different seasonings to find your favorite flavor combination.
### Timing for Cooking Crabs?
For best results, make sure you steam the crabs for about 15-20 minutes, or until they turn bright red. Steaming is a popular crab cooking technique because it helps retain the natural flavors and textures of the meat.
Here are a few key points to remember when cooking crabs:
• Choose fresh crabs that are alive and active.
• Clean the crabs thoroughly before cooking to remove any dirt or debris.
• Add spices and seasonings to the steaming water for extra flavor.
Steaming the crabs for the right amount of time is crucial to achieve perfectly cooked meat. Overcooking can result in rubbery and tough crab meat, while undercooking can leave the meat raw and unsafe to eat.
Once the crabs are cooked to perfection, it’s time to incorporate the delicious butter sauce to enhance their flavor and richness.
## Incorporating the Butter Sauce With the Crabs
Mix the butter sauce with the crabs to enhance their flavor and richness. The combination of the creamy, buttery sauce with the succulent meat of the crabs creates a mouthwatering experience that is hard to resist. The butter sauce adds a luxurious and indulgent touch to the already delicious crabs, elevating them to a whole new level. It coats each piece of crab meat in a velvety smoothness, enhancing its natural sweetness and bringing out its delicate flavors. The rich and savory sauce complements the sweetness of the crabs perfectly, creating a harmonious balance of flavors. Whether you are enjoying a simple crab boil or a fancy crab dish, incorporating the butter sauce will surely take your meal to the next level.
Pros Cons
Enhances flavor High in calories
Adds richness May be too heavy for some
Complements the sweetness of the crabs Can overpower delicate flavors
## Serving and Enjoying the Butter Sauce Crabs
To fully savor the delectable combination of butter sauce and crabs, simply dip a piece of succulent crab meat into the velvety, indulgent sauce and enjoy the explosion of flavors. The rich and creamy butter sauce enhances the natural sweetness of the crab meat, creating a mouthwatering experience.
Here are some serving suggestions and pairing options to elevate your enjoyment even further:
• Serve the butter sauce crabs with a side of warm, crusty bread to soak up every last drop of the sauce.
• Pair the dish with a chilled glass of crisp white wine, such as a Chardonnay or Sauvignon Blanc, to complement the buttery flavors.
• For a refreshing twist, serve the butter sauce crabs with a fresh salad tossed in a light vinaigrette dressing, providing a contrast to the richness of the dish.
With these serving suggestions and pairing options, you can enhance the experience of enjoying butter sauce crabs and create a memorable meal.
### Can I Use Margarine Instead of Butter for the Sauce?
Yes, you can use margarine instead of butter for the sauce. Margarine is a suitable substitute and will give a similar flavor. There are different types of butter sauces, but margarine can be used in most recipes.
### How Long Does It Take to Infuse the Flavors Into the Sauce?
Infusion time for flavor enhancement depends on personal preference. Longer infusion times, such as 30 minutes to an hour, allow the flavors to meld together more deeply, resulting in a richer, more complex sauce.
### Can I Use Frozen Crabs for This Recipe?
Yes, you can use frozen crabs for this recipe. While fresh crabs may provide a better taste, frozen crabs can still be delicious. Additionally, if you prefer, you can try alternative sauces for your crabs, like garlic butter or lemon herb.
### Can I Store the Leftover Butter Sauce for Later Use?
Yes, you can definitely store the leftover butter sauce for later use. Just make sure to transfer it to an airtight container and refrigerate it. When reheating, gently warm it in a saucepan over low heat, stirring occasionally until heated through.
### What Other Dishes Can I Serve the Butter Sauce With, Besides Crabs?
I love getting creative with butter sauce! It’s not just for crabs. You can drizzle it over grilled vegetables, dip steamed lobster in it, or even use it as a sauce for pasta. Delicious possibilities await!
## Conclusion
In conclusion, making butter sauce for crabs is a delightful culinary adventure that will elevate your crab feast to a whole new level.
While some may argue that butter sauce is high in calories and may not be the healthiest option, it’s important to remember that indulging in delicious flavors and enjoying the simple pleasure of a decadent meal is sometimes worth the extra calories.
So go ahead, treat yourself to a delectable butter sauce and savor the heavenly taste of perfectly cooked crabs. You deserve it!
# How to Make Butter Sauce for Lobster
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Making butter sauce for lobster is like adding a touch of velvety richness to an already exquisite dish.
In this article, I will guide you through the steps of creating a delectable butter sauce that perfectly complements the succulent flavors of lobster.
With a few simple ingredients and a little know-how, you’ll be able to elevate your lobster dining experience to new heights.
So, grab your apron and let’s get started on this culinary adventure!
## Key Takeaways
• Clarified butter is the best option for butter sauce for lobster.
• Experiment with adding herbs, garlic, or lemon zest to the butter for added flavor.
• Master prepping techniques like twisting and pulling the tail, cracking the claws, and cleaning the tomalley before cooking the lobster.
• Consider pairing lobster with crisp Chardonnay or sparkling Champagne, or experiment with different sauce variations like lemon-garlic or spicy tomato.
## Choosing the Right Butter
When making butter sauce for lobster, it’s important to choose the right butter. The best option for this sauce is clarified butter. Clarifying butter involves removing impurities such as milk solids and water, resulting in a rich and smooth texture. This process also increases the butter’s smoke point, making it ideal for cooking.
Additionally, using clarified butter enhances the flavor of the sauce as it allows the natural taste of the butter to shine through without any distractions. However, if you want to add an extra twist to your butter sauce, you can explore flavored butter options. Experiment with adding herbs, garlic, or lemon zest to the butter to infuse it with a burst of complementary flavors.
## Gathering Ingredients
First, you’ll need to gather all the necessary ingredients for your delicious butter sauce to accompany the lobster. Here’s a shopping list to help you out:
Ingredient Amount
Butter 1 cup
Garlic 3 cloves, minced
Lemon juice 1 tablespoon
Salt 1/2 teaspoon
Pepper 1/4 teaspoon
Parsley 1 tablespoon, chopped
If you don’t have some of these ingredients on hand, here are a few ingredient substitutions you can try:
• Instead of butter, you can use margarine or ghee.
• If you don’t have fresh garlic, you can use garlic powder or garlic paste.
• Lemon juice can be replaced with white wine vinegar or apple cider vinegar.
• Salt and pepper can be adjusted to taste, or you can use other seasonings like paprika or thyme.
Now that you have your shopping list and substitutions, you’re ready to gather the ingredients and start making your butter sauce for the lobster.
## Preparing the Lobster
When it comes to preparing and cooking lobster, there are a few key points to keep in mind.
First, understanding different lobster cooking techniques can help you achieve the perfect texture and flavor.
Additionally, knowing some lobster prepping tips can make the process easier and more efficient.
Lastly, exploring lobster flavor pairings can elevate the overall dining experience and bring out the best in this delicious seafood.
### Lobster Cooking Techniques
To achieve a tender lobster, it’s essential to master the cooking techniques.
Boiling lobsters is a common method that ensures a juicy and flavorful result. Start by bringing a pot of salted water to a rolling boil. Gently place the live lobsters into the pot and cover with a lid. Cook for about 8-10 minutes per pound.
Alternatively, grilling lobsters adds a smoky and charred flavor to the meat. Preheat the grill to medium-high heat and brush the lobsters with melted butter or oil. Place them shell-side down on the grill and cook for about 5-7 minutes per side.
Both boiling and grilling methods produce delicious lobsters, so choose the technique that suits your preference.
### Lobster Prepping Tips
One important tip for prepping lobsters is to remove the rubber bands from their claws. This allows the lobsters to move more freely and prevents any potential harm.
When it comes to lobster prepping, here are some key techniques to keep in mind:
• Cooking Methods:
• Boiling: A classic method that results in tender meat.
• Steaming: A healthier option that preserves the lobster’s natural flavors.
• Grilling: Adds a smoky, charred taste to the lobster.
• Shell Removal:
• Twist and Pull: Gently twist the tail and pull it away from the body.
• Crack and Break: Use a lobster cracker to carefully crack the claws and remove the meat.
• Clean the Tomalley: Remove the greenish substance found in the body cavity.
By mastering these prepping techniques, you’ll be well on your way to enjoying a delicious lobster dish.
Now, let’s move on to exploring some exciting lobster flavor pairings.
### Lobster Flavor Pairings
Let’s explore some exciting flavor pairings to enhance your lobster dish. When it comes to lobster, finding the perfect wine pairing can elevate the dining experience. Consider serving a crisp Chardonnay or a sparkling Champagne to complement the rich and buttery flavors of the lobster. If you prefer a red wine, opt for a light-bodied Pinot Noir. In addition to wine, experimenting with different sauce variations can add depth and complexity to your lobster dish. From classic butter sauce to zesty lemon-garlic sauce, there are endless possibilities to explore. To help you get started, here’s a table showcasing three delicious lobster sauce variations:
Sauce Variation Flavor Profile
Butter Sauce Creamy and indulgent
Lemon-Garlic Sauce Bright and tangy
Spicy Tomato Sauce Bold and fiery
Now that we’ve explored some exciting flavor pairings and sauce variations, let’s move on to making the base sauce for our lobster dish.
## Making the Base Sauce
Start by melting the butter in a saucepan over low heat. This is the base for your butter sauce, which will provide a creamy and rich flavor to complement the lobster. To achieve the perfect butter sauce consistency, make sure to melt the butter slowly to avoid burning it.
Once melted, you can add flavorful enhancements to customize the taste to your liking. Here are some options:
• Garlic: Saute minced garlic in the butter for a savory kick.
• Lemon: Squeeze fresh lemon juice into the sauce for a tangy twist.
• Herbs: Add chopped herbs like parsley, thyme, or chives for an aromatic touch.
By adding these flavorful enhancements to the melted butter, you can create a delicious and versatile sauce that will enhance the natural sweetness of the lobster.
Now, let’s move on to the next section and explore how to take your butter sauce to the next level.
Now that we have our base sauce ready, it’s time to take it up a notch and add some flavorful enhancements to enhance the taste of our butter sauce for lobster. Get ready to unleash your creativity with these creative twists!
One delicious way to enhance the flavor is by adding a hint of citrus. Squeeze in some fresh lemon juice to give a zesty kick to your sauce. Alternatively, you can experiment with other citrus fruits like orange or lime for a unique twist.
If you’re a fan of herbs, consider incorporating chopped fresh herbs like parsley, dill, or chives into the sauce. These herbs will not only add freshness but also provide a delightful aroma.
For a touch of richness and depth, a splash of white wine or brandy can do wonders. The alcohol will evaporate during cooking, leaving behind a subtle yet complex flavor.
With these flavorful enhancements, your butter sauce for lobster will truly shine with deliciousness. Let your taste buds guide you as you explore the endless possibilities to create a sauce that will elevate your lobster dish to new heights.
## Incorporating the Butter
Once the base sauce is ready, it’s time to incorporate the butter for a creamy and rich texture. To achieve the best results, I recommend using clarified butter, as it has a higher smoke point and imparts a clean, buttery flavor. However, if you prefer a more pronounced taste, you can also experiment with different types of butter in the sauce.
Here are three options to consider:
1. Salted Butter: This adds a subtle hint of saltiness, enhancing the overall flavor of the sauce.
2. Unsalted Butter: Using unsalted butter allows you to control the saltiness of the sauce, ideal for those who prefer a milder taste.
3. Cultured Butter: If you want to elevate the sauce with a tangy and nutty flavor, cultured butter is the way to go.
Remember to slowly whisk in the butter, allowing it to melt gradually and emulsify with the sauce. This will create a velvety smooth consistency that beautifully coats the lobster.
## Serving and Enjoying the Butter Sauce
To fully savor the velvety smooth consistency of the butter sauce, take a succulent bite of the perfectly cooked lobster, allowing the flavors to meld together in a tantalizing explosion on your palate.
When it comes to serving techniques for butter sauce, simplicity is key. Drizzle the warm sauce generously over the lobster, making sure to coat each piece evenly.
For an elegant presentation, serve the lobster on a bed of buttery mashed potatoes or alongside grilled asparagus. Alternatively, the butter sauce can be used as a dipping sauce for crusty bread or steamed vegetables.
Its rich and creamy flavor will elevate any dish it accompanies. So, whether you choose to pour it over your lobster or explore its alternative uses, this butter sauce will undoubtedly enhance your dining experience.
### How Can I Tell if the Lobster Is Cooked Properly Before Making the Butter Sauce?
To properly cook lobster, there are a few tips to determine if it’s cooked. Look for a bright red shell, opaque flesh, and an internal temperature of 145°F. These indicators ensure a delicious lobster for your butter sauce.
### Can I Use Margarine or a Different Type of Fat Instead of Butter for the Sauce?
Yes, you can use margarine or a different type of fat instead of butter for the sauce. However, it’s important to consider the pros and cons of each alternative fat to ensure the desired taste and texture are achieved.
### Is It Necessary to Use Fresh Lobster for This Recipe, or Can I Use Pre-Cooked Lobster?
Using pre-cooked lobster is a suitable alternative for this recipe. It saves time and effort. However, the taste and texture may differ slightly. For the best results, using fresh lobster is recommended.
### Can I Make the Butter Sauce Ahead of Time and Reheat It Later?
Yes, you can make the butter sauce ahead of time and reheat it later. It’s a convenient option for saving time. Also, there are various variations of butter sauce you can explore to add more flavor to your dish.
### Are There Any Alternative Spices or Herbs That Can Be Used to Enhance the Flavor of the Butter Sauce?
There are plenty of alternative ingredients and creative ways to enhance the flavor of butter sauce. Experimenting with different spices and herbs can add depth and complexity to the sauce, making it even more delicious.
## Conclusion
In conclusion, creating a decadent butter sauce for lobster is a culinary journey that takes your taste buds on a symphony of flavors. Don’t underestimate the power of choosing the right butter, it sets the foundation for a velvety and rich sauce.
By gathering the finest ingredients and preparing the lobster with care, you set the stage for a truly exceptional dish. Adding flavorful enhancements like herbs and spices elevates the sauce to new heights.
And finally, incorporating the butter brings everything together, creating a luscious and indulgent sauce that will leave you craving for more. So go ahead, serve up your masterpiece and savor every heavenly bite.
# How to Make Butter Sauce for Crab Legs
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I’ve discovered the perfect recipe for making butter sauce for crab legs, and I can’t wait to share it with you! This mouthwatering sauce will take your crab legs to a whole new level of deliciousness.
With just a few simple ingredients and easy steps, you’ll be able to create a creamy, flavorful sauce that complements the sweet, succulent meat of the crab perfectly.
Get ready to wow your taste buds with this indulgent and irresistible butter sauce. Let’s get cooking!
## Key Takeaways
• Melt the butter slowly over low heat and stir constantly to prevent burning
• Consider clarifying the butter for a smoother texture
• Add herbs, spices, and other flavorings to customize the butter sauce
• Serve the butter sauce generously over cooked crab legs and pair with lemon wedges and melted butter for dipping.
## Ingredients Needed
To make butter sauce for crab legs, you’ll need butter, garlic, lemon juice, and parsley.
There are different variations of butter sauce that you can try, depending on your personal preference. Some alternative ingredients you can use to enhance the flavor include white wine, Dijon mustard, or even a sprinkle of cayenne pepper for a touch of heat.
These ingredients can be easily found in your pantry or local grocery store. By experimenting with different variations and alternative ingredients, you can create a sauce that perfectly complements the succulent taste of crab legs.
Now that you have all the necessary ingredients, let’s move on to preparing the butter for the sauce.
## Preparing the Butter
When it comes to preparing the butter for my crab legs, there are a few key techniques that I always use to ensure the perfect consistency.
First, I like to melt the butter slowly over low heat, stirring constantly to prevent it from burning. This method allows the butter to melt evenly and smoothly, resulting in a velvety texture.
Additionally, I love to flavor the butter with some herbs and spices to enhance the overall taste. Adding a sprinkle of garlic powder, a pinch of paprika, and a squeeze of lemon juice adds a delicious depth of flavor to the butter that pairs perfectly with the sweet crab meat.
### Butter Melting Techniques
Melt the butter slowly to avoid it burning. It is important to use the right technique when melting butter for your crab legs sauce. Here are some tips to ensure success:
• Use a low heat setting on the stove to gradually melt the butter.
• Stir the butter constantly to distribute the heat evenly and prevent hotspots.
• Avoid using a microwave as it can cause the butter to splatter and unevenly melt.
• Consider clarifying the butter beforehand to remove any impurities and achieve a smoother texture.
• If you prefer a nutty flavor, you can also try browning the butter slightly for a richer taste.
By following these steps, you can achieve the perfect melted butter for your crab legs sauce.
Enjoy the delightful combination of flavors and indulge in a delicious seafood feast!
### Flavoring the Butter
Enhance the taste of your butter by adding herbs or spices. Flavoring techniques can elevate the simple butter sauce for crab legs into a culinary delight. Experimenting with flavors allows you to customize the butter to your liking and create a unique dining experience.
By infusing the butter with different herbs and spices, you can add complexity and depth to the sauce. For a hint of freshness, try incorporating chopped parsley or dill. If you prefer a little heat, add a pinch of cayenne pepper or a dash of hot sauce. The possibilities are endless, so don’t be afraid to get creative and try different combinations.
Now that you know about flavoring techniques, let’s move on to selecting the right herbs and spices to enhance your butter sauce even further.
## Selecting the Right Herbs and Spices
To create the perfect butter sauce for your crab legs, start by choosing the right herbs and spices. Experiment with different herb combinations to elevate the flavor profile of your sauce. Some popular herb options to consider are fresh dill, parsley, chives, tarragon, and basil. Each herb adds its own unique taste, from tangy and refreshing to sweet and earthy.
In addition to herbs, exploring alternative spice options can bring a new dimension to your butter sauce. Try spices like paprika, cayenne pepper, or even a touch of curry powder to add some heat and complexity. Remember to balance the flavors and find the combination that suits your taste preferences.
Don’t be afraid to get creative and experiment with different herb and spice combinations until you find your perfect butter sauce for crab legs.
## Cooking the Crab Legs
When it comes to cooking crab legs, two key factors to consider are the cooking time and temperature. It’s important to ensure that the crab legs are cooked thoroughly but not overcooked, resulting in a tough and rubbery texture.
Additionally, seasonings and flavor options play a crucial role in enhancing the taste of the crab legs, allowing you to customize the dish to your preferences and create a delicious dining experience.
### Cooking Time and Temperature
The butter sauce is ready once it’s heated to the desired temperature. Cooking techniques and knowing the optimal temperature are crucial for achieving a perfect butter sauce for your crab legs.
Here are some key points to keep in mind:
• Butter Melting: Start by melting the butter in a saucepan over low heat. This allows the butter to slowly melt without burning.
• Simmering: Once melted, increase the heat to medium-low and let the butter simmer gently. This helps to infuse the flavors.
• Avoid Boiling: Be careful not to let the butter sauce boil, as it can cause separation and a greasy texture.
• Whisking: Whisk the butter sauce constantly to ensure a smooth and creamy consistency.
• Seasoning: Add your desired seasonings, such as garlic, lemon juice, or herbs, to enhance the flavor of the sauce.
### Seasonings and Flavor Options
Now that we know the perfect cooking time and temperature for our crab legs, let’s talk about the seasonings and flavor options for our butter sauce. There are endless possibilities when it comes to flavor pairing ideas, so you can get creative and experiment with different combinations. To give you some inspiration, here are a few popular options:
Flavor Pairing
Garlic Lemon
Old Bay seasoning Cajun spice
Dill Mustard
In addition to the traditional boiling method, you can also try alternative cooking methods like grilling or steaming the crab legs for a different twist. These methods can add a smoky or delicate flavor to the crab meat, enhancing the overall taste of the dish.
Now that we have some flavor pairing ideas and alternative cooking methods, let’s move on to preparing the base for our mouthwatering butter sauce.
## Preparing the Base for the Sauce
To make butter sauce for crab legs, start by melting the butter in a saucepan. This step is crucial as it allows the butter to become a smooth and creamy base for the sauce.
As the butter melts, it releases its rich aroma, infusing flavors into the sauce. The tantalizing scent of melted butter fills the kitchen, creating a mouthwatering anticipation for the delicious sauce to come.
The butter slowly transforms from solid to liquid, enhancing its aroma and adding a velvety texture to the sauce. As the butter melts, it becomes ready to be combined with other ingredients, further enriching the flavors of the sauce.
Once the butter has completely melted, it is time to move on to the next step of adding the butter to the sauce, bringing us one step closer to a delectable butter sauce for our crab legs.
## Adding the Butter to the Sauce
Start by melting the butter in a saucepan, allowing it to become a smooth and creamy base for the sauce.
Butter is a classic choice for a sauce, but there are also alternative options available. For those who prefer a lighter sauce, olive oil can be used instead. It adds a delicate flavor that complements the sweetness of the crab legs.
Another alternative is using a combination of butter and lemon juice, which adds a tangy twist to the sauce.
When it comes to storing leftover butter sauce, it’s important to refrigerate it in an airtight container. This will help maintain its freshness and prevent any contamination.
Leftover butter sauce can be kept in the refrigerator for up to 3 days, but it’s always best to use it as soon as possible for optimal flavor.
## Incorporating the Herbs and Spices
Enhance the flavor of your sauce by incorporating a variety of herbs and spices such as garlic, thyme, and paprika. Herbs and spices are essential in elevating the taste of any dish, and butter sauce for crab legs is no exception.
When selecting the right herbs and spices for your sauce, consider the following factors:
• Freshness: Choose herbs and spices that are fresh to ensure optimal flavor.
• Compatibility: Consider the flavors that pair well with seafood, like dill or parsley.
• Intensity: Adjust the amount of herbs and spices based on your personal preference.
• Balance: Aim for a balanced combination of flavors, avoiding overpowering or clashing tastes.
• Experimentation: Don’t be afraid to try new combinations to find your perfect blend.
## Serving and Enjoying the Butter Sauce
Once you’ve prepared your delectable butter sauce, it’s time to savor its rich flavors by drizzling it generously over your perfectly cooked crab legs.
The butter sauce adds a creamy and savory element to the sweet and succulent crab meat, enhancing its natural flavors.
When serving the butter sauce, it’s a good idea to have some lemon wedges on the side. A squeeze of fresh lemon juice can brighten up the dish and cut through the richness of the butter.
Additionally, you can serve the crab legs with a side of melted butter for dipping, allowing your guests to indulge in even more buttery goodness.
As for pairing options, a crisp and refreshing white wine, like a Chardonnay or Sauvignon Blanc, complements the flavors of the crab legs and butter sauce perfectly. Enjoy!
### How Long Should I Cook the Crab Legs Before Adding the Butter Sauce?
I typically cook the crab legs for about 10-12 minutes before adding the butter sauce. This allows the crab legs to fully cook and ensures that the butter sauce will have the perfect consistency.
### Can I Use Margarine Instead of Butter for the Sauce?
Sure, you can use margarine instead of butter for the sauce. However, keep in mind that margarine has a slightly different taste compared to butter. It’s a matter of personal preference.
### Is It Necessary to Use Fresh Herbs, or Can I Use Dried Herbs Instead?
Using dried herbs in butter sauce is a matter of personal preference. Fresh herbs offer a vibrant flavor, while dried herbs provide convenience and longer shelf life. Experiment and find what suits your taste buds.
### Can I Refrigerate the Leftover Butter Sauce and Reheat It Later?
Yes, you can refrigerate the leftover butter sauce and reheat it later. It’s important to store it in an airtight container and gently reheat it on low heat to prevent separation.
### What Are Some Alternative Ways to Enjoy the Butter Sauce, Besides With Crab Legs?
There are countless creative ways to incorporate butter sauce in recipes! It’s not just for crab legs. It adds a rich and indulgent touch to dishes like grilled vegetables, pasta, and even roasted chicken. The possibilities are endless!
## Conclusion
And there you have it, a delectable butter sauce for your crab legs that will take your dining experience to the next level.
The rich and creamy texture of the butter perfectly complements the succulent meat of the crab, creating a harmonious blend of flavors.
With the right combination of herbs and spices, this sauce becomes a tantalizing indulgence that will leave you craving for more.
So go ahead, treat yourself to this culinary delight and savor every mouthful of buttery goodness.
Bon appétit!
#### Affiliate disclaimer
As an affiliate, we may earn a commission from qualifying purchases. We get commissions for purchases made through links on this website from Amazon and other third parties.
#### Trending
Copyright © 2017 Eat More Butter Affiliate disclaimer As an affiliate, we may earn a commission from qualifying purchases. We get commissions for purchases made through links on this website from Amazon and other third parties. | 9,252 | 44,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-50 | longest | en | 0.907239 |
https://numberworld.info/12712101 | 1,603,686,077,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890273.42/warc/CC-MAIN-20201026031408-20201026061408-00311.warc.gz | 440,817,520 | 3,879 | Number 12712101
Properties of number 12712101
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
c1f8a5
Base 32:
c3u55
sin(12712101)
-0.97048136293021
cos(12712101)
-0.24117612693865
tan(12712101)
4.0239528482729
ln(12712101)
16.358064932416
lg(12712101)
7.1042173347632
sqrt(12712101)
3565.4033432418
Square(12712101)
Number Look Up
Look Up
12712101 which is pronounced (twelve million seven hundred twelve thousand one hundred one) is a very great number. The cross sum of 12712101 is 15. If you factorisate the figure 12712101 you will get these result 3 * 401 * 10567. The figure 12712101 has 8 divisors ( 1, 3, 401, 1203, 10567, 31701, 4237367, 12712101 ) whith a sum of 16993344. 12712101 is not a prime number. The number 12712101 is not a fibonacci number. The figure 12712101 is not a Bell Number. The figure 12712101 is not a Catalan Number. The convertion of 12712101 to base 2 (Binary) is 110000011111100010100101. The convertion of 12712101 to base 3 (Ternary) is 212220211201120. The convertion of 12712101 to base 4 (Quaternary) is 300133202211. The convertion of 12712101 to base 5 (Quintal) is 11223241401. The convertion of 12712101 to base 8 (Octal) is 60374245. The convertion of 12712101 to base 16 (Hexadecimal) is c1f8a5. The convertion of 12712101 to base 32 is c3u55. The sine of 12712101 is -0.97048136293021. The cosine of 12712101 is -0.24117612693865. The tangent of the figure 12712101 is 4.0239528482729. The square root of 12712101 is 3565.4033432418.
If you square 12712101 you will get the following result 161597511834201. The natural logarithm of 12712101 is 16.358064932416 and the decimal logarithm is 7.1042173347632. I hope that you now know that 12712101 is unique number! | 671 | 1,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-45 | latest | en | 0.782535 |
http://math.stackexchange.com/questions/350857/reasoning-about-the-gamma-function-using-the-digamma-function | 1,469,613,202,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826759.85/warc/CC-MAIN-20160723071026-00109-ip-10-185-27-174.ec2.internal.warc.gz | 161,725,296 | 17,467 | # Reasoning about the gamma function using the digamma function
I am working on evaluating the following equation:
$\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)$
If I'm understanding correctly, the above is an increasing function which can be demonstrated by the following argument using the digamma function $\frac{\Gamma'}{\Gamma}(x) = \int_0^\infty(\frac{e^{-t}}{t} - \frac{e^{-xt}}{1-e^{-t}})$:
$\frac{\Gamma'}{\Gamma}(\frac{1}{2}x) - \frac{\Gamma'}{\Gamma'}(\frac{1}{3}x) = \int_0^\infty\frac{1}{1-e^{-t}}(e^{-\frac{1}{3}xt} - e^{-\frac{1}{2}xt})dt > 0 (x > 1)$
Please let me know if this reasoning is incorrect or if you have any corrections.
Thanks very much!
-Larry
-
The series for $\psi$ might be more expedient to use... – J. M. Apr 4 '13 at 6:33
Thanks very much! I'll check it out. – Larry Freeman Apr 4 '13 at 6:51
I reviewed the series for $\psi$. Thanks. Is this correct: Using $\psi(x) = -\gamma+\sum_{k=0}^\infty(\frac{1}{k+1}-\frac{1}{k+z})$ gets me to the derivative of $\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)$ to be: $\sum_{k=0}^{\infty}(\frac{1}{k+\frac{x}{3}} - \frac{1}{k+\frac{x}{2}})$ which shows an increasing function. – Larry Freeman Apr 4 '13 at 7:04
Well, the terms of your resulting series are all positive for positive argument, so... – J. M. Apr 4 '13 at 11:05
Great. Thanks for the help. – Larry Freeman Apr 4 '13 at 14:21
This answer is provided with help from J.M.
$\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)$ is an increasing function. This can be shown using this series for $\psi$:
The function is increasing if we can show: $\frac{d}{dx}(\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)) > 0$
We can show this using the digamma function $\psi(x)$:
$$\frac{d}{dx}(\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)) = \frac{\psi(\frac{1}{2}x)}{2} - \frac{\psi(\frac{1}{3}x)}{3}$$
$$\frac{\psi(\frac{1}{2}x)}{2} - \frac{\psi(\frac{1}{3}x)}{3} = -\gamma + \sum_{k=0}^\infty(\frac{1}{k+1} - \frac{1}{k + {\frac{1}{2}}}) + \gamma - \sum_{k=0}^\infty(\frac{1}{k+1} - \frac{1}{k+\frac{1}{3}})$$
$$= \sum_{k=0}^\infty(\frac{1}{k+\frac{1}{3}} - \frac{1}{k+\frac{1}{2}})$$
Since for all $k\ge 0$: $k + \frac{1}{3} < k + \frac{1}{2}$, it follows that for all $k\ge0$: $\frac{1}{k+\frac{1}{3}} > \frac{1}{k+\frac{1}{2}}$ and therefore: $$\sum_{k=0}^\infty(\frac{1}{k+\frac{1}{3}} - \frac{1}{k+\frac{1}{2}}) > 0.$$
- | 950 | 2,394 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2016-30 | latest | en | 0.645779 |
http://slideplayer.com/slide/4101559/ | 1,506,127,706,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689411.82/warc/CC-MAIN-20170922235700-20170923015700-00077.warc.gz | 316,815,448 | 22,874 | # Body Mass Index 101 It’s like paint by numbers for grownups!
## Presentation on theme: "Body Mass Index 101 It’s like paint by numbers for grownups!"— Presentation transcript:
Body Mass Index 101 It’s like paint by numbers for grownups!
Fun Facts about BMI Well… fun if you like body weight math
What is BMI? A little knowledge is actually a great thing BMI stands for body mass index. It’s a measure of your weight as it relates to your height. Knowing your BMI will help you determine whether or not you face various health risks.
How Do You Calculate BMI? First, take your weight in pounds and multiply it by 703. Second, multiply your height in inches by itself. Then divide the first number by the second number! How easy is that?
For Example… If a man is 70 inches tall (5’ 10’’) and weighs 240 pounds, the BMI math should look like this… (240 x 703) / (70 x 70) = BMI (168,720) / (4,900) = BMI 34.43 = BMI
Online: BMI Calculator
BMI Guidelines Keep your friends close and your measurements closer Normal BMI is 18.5 to 24.9. Overweight BMI is 25 to 29.9. Obese BMI is 30 or above. |||||||||||||||||| 1520253035 Normal OverweightObese
But Wait…There’s More! Man cannot live by BMI alone |||||||||||||||||| 1520253035
BMI Caveats BMI is not a perfect measure of obesity. –If you’re especially muscular, your BMI result may overestimate your body fat. –BMI measurements are less accurate among people who are 5’ tall or shorter. –BMI results are less accurate among the elderly. BMI is not a perfect measure of obesity. –If you’re especially muscular, your BMI result may overestimate your body fat. –BMI measurements are less accurate among people who are 5’ tall or shorter. –BMI results are less accurate among the elderly.
BMI and Waist Circumference Just the facts man, just the facts If you carry your weight around your waist, you may be at additional risk for obesity-associated complications--even if your BMI is relatively low. To find out if you’re at risk, measure the circumference of your waist.
Learn to do it! To measure your waist circumference, stand up, wrap a tape measure around your waist (just above your hips), and breathe out. Once you breathe out, look at the tape measure and record the number of inches it takes to circle your waist.
Waist Circumference When to Worry For men, a waist circumference of 40 inches or more. For women, that magic number is 35 inches or more.
Quick Quiz True or False: If a woman’s waist circumference is 35 inches or more, she is at greater risk of developing heart disease or diabetes. You are obese if your BMI is over: * 25 * 35 * 30* 40 True or False: If a woman’s waist circumference is 35 inches or more, she is at greater risk of developing heart disease or diabetes. You are obese if your BMI is over: * 25 * 35 * 30* 40
BMI, Waist Circumference, and You One big, happy family
Risky Business People with high BMI or waist circumference increase their risk of getting… –Type 2 diabetes-- Heart disease –High blood pressure-- Stroke –Liver disease-- Sleep apnea –Osteoarthritis-- Gallbladder disease
Contributing Factors All rest and no exercise makes Jack an unhealthy boy If you’re a smoker,don’t exercise much,have high cholesterol, or eat unhealthy food, having a high BMI or waist circumference significantly increases your risk of getting heart disease or type 2 diabetes.
Being Overweight or Obese is Dangerous According to the National Institute of Health, over 300,000 lives could be saved annually if people maintained a healthy weight. When your body has to carry so much excess weight, it puts tremendous strain on your heart.
Quick Quiz True or False: If your BMI is high, but you also have high cholesterol, you decrease your risk of heart disease. People with high BMI or waist circumference are at risk of getting… * Type 2 diabetes * Heart disease * High blood pressure* All of the above True or False: If your BMI is high, but you also have high cholesterol, you decrease your risk of heart disease. People with high BMI or waist circumference are at risk of getting… * Type 2 diabetes * Heart disease * High blood pressure* All of the above
What Can You Do to Lower Your BMI or Waist Circumference? It’s as easy as pie. Well, as easy as saying “no” to pie.
Change What You Can Since it’s very difficult to adjust your height, the way to lower your BMI is by losing weight. Losing weight will also lower your waist circumference.
…But How? Changing diet and lifestyle is easier said than done
Where to Start Eat more fruits, vegetables, beans and fiber-rich foods. Eat less fat, sugar and sodium. You don’t have to run a marathon to begin an exercise plan. Walking is a great way to start.
Set Achievable Goals Your goals should be specific and attainable. Saying “I should lose weight” is a great first step, but follow it up with a plan. For example, walking for 30 minutes, 5 times a week is a specific, attainable initial goal.
Set Up Rewards for Doing Well Just in case virtue isn’t enough Reward yourself for making progress along the way, not just for reaching your overall goal. THE REWARD SHOULD NOT BE FOOD!
Adjusting Rewards To the victor go small, frequent shares of the spoils Earning frequent, small rewards for achieving smaller goals is a more effective way to change habits than celebrating a long, hard effort with a large reward.
Hold Yourself Accountable Keep a journal that records your daily progress and setbacks. If you’re changing your diet, record what you eat each day. If you’re increasing exercise, keep a list of what you do and when.
Keeping Track of Progress To err is human. To record that error in your journal is divine. Reflect on your progress--these notes could help you make connections you might not be aware of.
Words of Hope! Losing as little as 5- 7% of your body weight significantly decreases your health risks! A fantastic first step is to lose 10% of your body weight. You can do it! Your heart will thank you.
Quick Quiz True or false: It’s a great idea to reward your diet and exercise successes with food. Set ___, ___ goals * Large, achievable* Specific, attainable * General, difficult* Specific, difficult True or false: It’s a great idea to reward your diet and exercise successes with food. Set ___, ___ goals * Large, achievable* Specific, attainable * General, difficult* Specific, difficult
Now Let’s Test Your Memory! BMI stands for… * Body Mass Index* Body Mass Increase * Burly Man I.D.* Big Muscular Insect True or false: You measure waist circumference by wrapping a tape measure around your waist and recording the result. BMI stands for… * Body Mass Index* Body Mass Increase * Burly Man I.D.* Big Muscular Insect True or false: You measure waist circumference by wrapping a tape measure around your waist and recording the result.
That’s All, Folks! Do you have any questions?
Download ppt "Body Mass Index 101 It’s like paint by numbers for grownups!"
Similar presentations | 1,578 | 6,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-39 | longest | en | 0.858596 |
https://www.qb365.in/materials/stateboard/7th-standard-maths-t3-percentage-and-simple-interest-english-medium-free-online-test-1-mark-questions-with-answer-key-2020-2021-9556.html | 1,642,727,429,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00700.warc.gz | 992,431,554 | 31,215 | #### T3 - Percentage and Simple Interest English Medium Free Online Test 1 Mark Questions with Answer Key 2020 - 2021
7th Standard
Reg.No. :
•
•
•
•
•
•
Maths
Time : 00:10:00 Hrs
Total Marks : 9
Part A
9 x 1 = 9
1. Thendral saved one fourth of her salary. Her savings percentage is
(a)
$\cfrac { 3 }{ 4 }$%
(b)
$\cfrac { 1 }{ 4 }$%
(c)
25%
(d)
1%
2. Kavin scored 15 out of 25 in a test. The percentage of his marks is
(a)
60%
(b)
15%
(c)
25%
(d)
15/25
3. 0.07% is
(a)
$\cfrac { 7 }{ 10 }$
(b)
$\cfrac { 7 }{ 100 }$
(c)
$\cfrac { 7 }{ 1000 }$
(d)
$\cfrac { 7 }{ 10,000 }$
4. Decimal value of 142.5% is
(a)
1.425
(b)
0.1425
(c)
142.5
(d)
14.25
5. The percentage of 0.005 is
(a)
0.005%
(b)
5%
(c)
0.5%
(d)
0.05%
6. The percentage of 4.7 is
(a)
0.47%
(b)
4.7%
(c)
47%
(d)
470%
7. The interest for a principle of Rs.4,500 which gives an amount of Rs.5,000 at end of certain period is
(a)
Rs.500
(b)
Rs.200
(c)
Rs.20%
(d)
Rs.15%
8. Which among the following is the simple interest for the principle of ₹ 1,000 for one year at the rate of 10% interest per annum?
(a)
Rs.200
(b)
Rs.10
(c)
Rs.100
(d)
Rs.1,000
9. Which among the following rate of interest yields an interest of Rs.200 for the principle of Rs. 2,000 for one year.
(a)
10%
(b)
20%
(c)
5%
(d)
15% | 502 | 1,337 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-05 | latest | en | 0.768907 |
https://stats.stackexchange.com/questions/78340/modeling-prior-probability-as-a-delta-function?noredirect=1 | 1,723,271,461,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00684.warc.gz | 415,376,704 | 37,267 | # Modeling prior probability as a delta function [closed]
I'm using approximate Bayesian computation to find the true value of a parameter. My prior distribution is uniform over $(0, 1)$.
I was watching this video on Bayesian learning and the lecturer states (around 36:00) that this is making a huge assumption. That is, we are assuming the value of the unknown variable has the same mean as the random variable (in my case $0.5$) and also has a variance that we can compute (I assume based off the number of observations?). He goes on to say that if you want to really model a prior like this, you should use a delta function centered around a value, $a$, which is unknown.
My questions are these:
1. What is wrong with assuming that the mean and variance are the same as the random variable?
2. How can I use a delta function to generate probabilities for my prior?
• Not sure that I understand everything, but I don't see how you can use a delta function as a prior: it is like fixing your parameter at $a$. One possible problem is that the mean of your prior is equal to the true parameter, so if your likelihood has no information about the parameter, the posterior mean will be very close to the truth. So you might end up thinking that your estimation procedure is working well, when actually you are just sampling the prior. Commented Dec 3, 2013 at 13:38
• Also: unless I have misunderstood the lecturer, he is not saying that you should substitute the prior with a delta function. He is saying that the belief "my parameter takes a certain value, which I don't know" can be translated into a delta function centred around an unknown point $a$. In practice you can use a proper prior (such as your uniform), but you have to be aware that this is not equivalent to the above statement of ignorance, but you are making some assumptions. Commented Dec 3, 2013 at 13:50
• The question and the comment remain impenetrable to me... Commented Nov 19, 2015 at 20:06
• While I had thought that the question "can the delta function be used as a prior" made sense given (for example) the approach presented by Pishro-Nik especially for mixed random variables, I probably was "out of my depth" and have accordingly deleted my post here. Nobody cares and I couldn't care less. ;-)
– gwr
Commented Nov 27, 2015 at 12:13
• ohblahitsme: As Matteo said, a point mass (Dirac delta) prior says "I know the value of the parameter" -- in which case the data have no impact. Clearly given what you say the lecturer said about the other prior, a point prior cannot have been the intended prior. One is then left to wonder what is really meant. Can you clarify? Commented Nov 29, 2015 at 23:59 | 628 | 2,683 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-33 | latest | en | 0.950556 |
https://wiki.beyondunreal.com/Legacy:MilkShape_For_UT2003_Static_Meshes/Modeling_And_Smoothing_In_MS3D | 1,550,326,356,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480472.38/warc/CC-MAIN-20190216125709-20190216151709-00450.warc.gz | 732,785,769 | 9,688 | I search for solutions in this order: Past Code, Unreal Source, Wiki, BUF, groups.yahoo, google, screaming at monitor. – RegularX
# Legacy:MilkShape For UT2003 Static Meshes/Modeling And Smoothing In MS3D
NOTE: The pictures in this tutorial do not show the interface elements to relative scale.
### Modeling the shuriken
Launch MilkShape. Enable the Model tab. Enable the Cylinder tool button, enter Stacks = 4, Slices = 15 in the data entry fields. Close Cylinder should be checked.
In the Front viewport, place the pointer at X = -40, Y = 20, click LMB & drag to X = 40, Y = -20.
In the Tools box, enable Select. In the Select Options box, enable Vertex.
In the Front viewport, drag-select the vertices of the top surface (cap) of the cylinder (drag-select = LMB click & drag to lasso the vertices).
Add the vertices of the bottom cap to the selection (Shift + LMB drag-select adds vertices to the selection).
In the toolbox, enable Scale. In Scale Options, disable the Y axis button. In the data entry fields, enter .30 in X and Z (.30 = 30% scale factor). Click the Scale button. The result is shown in FIG. 1
Enable Select (shortcut key = F1), Options = Vertex. In the Top viewport, drag-select the vertices on the circumference of the cylinder that are located at the 3 o'clock position. Using this position as 1, count vertices around the circumference (either direction) and add the vertices at counts 6 and 11 to the selection.
Scale (shortcut key = F4) the selected vertices .50 in X & Z (FIG. 2).
Counting from position 1 on the circumference, select vertices at 3, 8 and 13. Scale the vertices 1.5 in X & Z (FIG. 3).
Enable the Rotate tool (shortcut key = F3). Using the data entry fields, rotate the selected vertices -10 in Y (FIG. 4).
Counting from position 1 on the circumference, select vertices at 5, 10 and 15. Scale the vertices .73 in X & Z. Rotate the vertices 10 in Y.
Counting from position 1 on the circumference, select vertices at 2, 7 and 12. Rotate the vertices -10 in Y.
Select vertices at 3, 8 and 13. With Rotate enabled, click LMB & drag in the Top viewport to rotate vertices slightly counterclockwise. In like manner, select, rotate, and scale sets of vertices at every 5th count, until the top view of the model closely resembles FIG. 5.
In the Left viewport, drag-select the vertices highlighted (red) in FIG. 6 and scale .50 in X & Z. Select the vertices of the cylinder caps and scale .73 in X & Z (FIG. 6).
Enable Select, Select Options=Face, and By Vertex is checked ( Select|Face|By Vertex ). Drag-select the top & bottom cap vertices. The faces connected to these vertices will be highlighted (FIG. 7A).
Enable the Groups tab. Click the Regroup button. Rename Regroup 1 to hubs (FIG. 7B).
Enable the Model tab. Enable Select|Face|By Vertex. In the Front viewport, drag-select the top two rows of vertices. The faces of the top half of the model will be highlighted.
Drag-deselect the vertices of the top cap (drag-deselect = Shift + click & drag with RMB). The faces just above the horizontal centerline of the model should be highlighted (FIG. 8A). Enable Groups, Regroup the faces, and rename them UpSurf (FIG. 8B).
Enable Model|Select|Face|By Vertex. As above, select the faces just below the model's horizontal centerline. Switch to the Groups tab, Regroup these faces and rename them DownSurf (FIG 9).
Click on the background of any ortho viewport (Front, Side or Top) to deselect all, or use the Shift+Ctrl+A shortcut.
In the Groups list, highlight hubs and click the Select button in the Group box. Faces in the hubs group will be highlighted.
In the Smoothing Groups box, enable the Assign button & click on the square smoothing group button labeled 1. In the 3D viewport, the faces of group hubs will be smoothed.
Deselect all, then highlight UpSurf in the Groups list & click Group|Select. In Smoothing Groups, assign smoothing group number 2 to UpSurf.
Deselect all, then highlight DownSurf in the Groups list & click Group|Select. In Smoothing Groups, assign smoothing group number 3 to DownSurf.
There should now be hard (unsmoothed) edges between hubs, UpSurf and DownSurf.
Switch to Model, select all (Ctrl+A), and scale the entire model .30 in Y only (disable the X & Z axis buttons).
Enable Select|Vertex and in the Front viewport drag-select the vertices of the caps, then Scale then .73 in X & Z.
In the Top viewport, drag-select the vertices at the origin (center of hubs). Enable Scale in Y only, then LMB click + drag in the Front viewport to scale up the central vertices a small amount in Y, rounding the off the hubs slightly. The modeling is complete. | 1,152 | 4,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-09 | latest | en | 0.78954 |
https://docs.bentley.com/LiveContent/web/STAAD.Pro%20Help-v15/ja/GUID-4493E464-E1CB-4EDE-ABBF-F1FAD5C4A276.html | 1,656,142,552,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034877.9/warc/CC-MAIN-20220625065404-20220625095404-00497.warc.gz | 257,000,834 | 5,023 | # G.15.3 Area, One-way, and Floor Loads
Often a floor is subjected to a uniform pressure. It could require a lot of work to calculate the equivalent member load for individual members in that floor. However, with the AREA, ONEWAY or FLOOR LOAD facilities, you can specify the pressure (load per unit square area). The program will calculate the tributary area for these members and calculate the appropriate member loads. The Area Load and Oneway load are used for one way distribution and the Floor Load is used for two way distribution.
1. The member load is assumed to be a linearly varying load for which the start and the end values may be of different magnitude.
2. Tributary area of a member with an area load is calculated based on half the spacing to the nearest approximately parallel members on both sides. If the spacing is more than or equal to the length of the member, the area load will be ignored. Oneway load does not have this limitation.
3. These loading types should not be specified on members declared as MEMBER CABLE, MEMBER TRUSS, MEMBER TENSION, MEMBER COMPRESSION, or CURVED.
Note: Floor Loads and One-way Loads can be reduced when included in a load case defined as "Reducible" according to the UBC/IBC rules.
An example:
### Example floor structure with area load specification of 0.1
Member 1 will have a linear load of 0.3 at one end and 0.2 at the other end. Members 2 and 4 will have a uniform load of 0.5 over the full length. Member 3 will have a linear load of 0.45 and 0.55 at respective ends. Member 5 will have a uniform load of 0.25. The rest of the members, 6 through 13, will have no contributory area load since the nearest parallel members are more than each of the member lengths apart. However, the reactions from the members to the girder will be considered.
Only member loads are generated from the Area, Oneway, and Floor load input. Thus, load types specific to plates, solids or surface are not generated. That is because, the basic assumption is that, a floor load or area load is used in situations where the basic entity (plate, solid or surface) which acts as the medium for application of that load, is not part of the structural model.
The Oneway load is intended to be used in areas with relatively large aspect rations between adjacent sides. It should not be used on members with square tributary areas unless the TOWARDS option is used, which then specifies to which of the two equal directions the load should be applied. Otherwise, the Floor Load type should be used.
Note: Failure to specify a TOWARD side on a Oneway load applied to a square tributary area will likely result in lost load or unintended load path changes. | 611 | 2,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-27 | latest | en | 0.951492 |
https://leetcode.ca/2021-10-11-2027-Minimum-Moves-to-Convert-String/ | 1,726,527,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00399.warc.gz | 312,986,980 | 7,195 | Formatted question description: https://leetcode.ca/all/2027.html
2027. Minimum Moves to Convert String (Easy)
You are given a string s consisting of n characters which are either 'X' or 'O'.
A move is defined as selecting three consecutive characters of s and converting them to 'O'. Note that if a move is applied to the character 'O', it will stay the same.
Return the minimum number of moves required so that all the characters of s are converted to 'O'.
Example 1:
Input: s = "XXX"
Output: 1
Explanation: XXX -> OOO
We select all the 3 characters and convert them in one move.
Example 2:
Input: s = "XXOX"
Output: 2
Explanation: XXOX -> OOOX -> OOOO
We select the first 3 characters in the first move, and convert them to 'O'.
Then we select the last 3 characters and convert them so that the final string contains all 'O's.
Example 3:
Input: s = "OOOO"
Output: 0
Explanation: There are no 'X's in s to convert.
Constraints:
• 3 <= s.length <= 1000
• s[i] is either 'X' or 'O'.
Solution 1.
• class Solution {
public int minimumMoves(String s) {
int ans = 0;
for (int i = 0; i < s.length(); ++i) {
if (s.charAt(i) == 'X') {
++ans;
i += 2;
}
}
return ans;
}
}
• class Solution {
public:
int minimumMoves(string s) {
int ans = 0;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == 'X') {
++ans;
i += 2;
}
}
return ans;
}
};
• class Solution:
def minimumMoves(self, s: str) -> int:
ans = i = 0
while i < len(s):
if s[i] == "X":
ans += 1
i += 3
else:
i += 1
return ans
• func minimumMoves(s string) (ans int) {
for i := 0; i < len(s); i++ {
if s[i] == 'X' {
ans++
i += 2
}
}
return
}
• function minimumMoves(s: string): number {
const n = s.length;
let ans = 0;
let i = 0;
while (i < n) {
if (s[i] === 'X') {
ans++;
i += 3;
} else {
i++;
}
}
return ans;
}
• impl Solution {
pub fn minimum_moves(s: String) -> i32 {
let s = s.as_bytes();
let n = s.len();
let mut ans = 0;
let mut i = 0;
while i < n {
if s[i] == b'X' {
ans += 1;
i += 3;
} else {
i += 1;
}
}
ans
}
} | 642 | 1,985 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-38 | latest | en | 0.600991 |
https://hollows.info/what-are-corresponding-angles-theorem/ | 1,702,055,049,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100762.64/warc/CC-MAIN-20231208144732-20231208174732-00723.warc.gz | 329,758,306 | 9,555 | ## What are corresponding angles Theorem?
The Corresponding Angles Postulate states that, when two parallel lines are cut by a transversal , the resulting corresponding angles are congruent . The converse is also true; that is, if two lines l and m are cut by a transversal in such a way that the corresponding angles formed are congruent , then l∥m .
## What are corresponding angles examples?
Definition: Corresponding angles are the angles which are formed in matching corners or corresponding corners with the transversal when two parallel lines are intersected by any other line (i.e. the transversal). For example, in the below-given figure, angle p and angle w are the corresponding angles.
What is the meaning of corresponding angles are corresponding sides?
Corresponding sides touch the same two angle pairs. When the sides are corresponding it means to go from one triangle to another you can multiply each side by the same number. In the diagram of similar triangles the corresponding sides are the same color.
### What are corresponding angles class 8?
Answer: Corresponding angles : If the arms on the transversal of a pair of angles are in the same direction and the other arms are on the same side of transversal, then it is called a pair of corresponding angles.
### How do you know if two angles are corresponding?
Corresponding angles are equal if the transversal intersects two parallel lines. If the transversal intersects non-parallel lines, the corresponding angles formed are not congruent and are not related in any way.
What are the properties of corresponding angles?
Corresponding Angles If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.
Alternate Exterior Angles Converse If two lines are cut by a transversal and the alternate exterior angles are congruent, the lines are parallel.
## Are corresponding angles always equal?
A pair of corresponding angles lie on the same side of the transversal. The corresponding pair of angles comprises one exterior angle and another interior angle. Not all corresponding angles are equal. Corresponding angles are equal if the transversal intersects two parallel lines.
## Where do corresponding angles lie?
Corresponding angles lie on the same side of a transversal and on the same side of two lines that are intersected by the transversal. Postulate 3.1 Corresponding Angles (p. 180) If two parallel lines are cut by a transversal, then each pair of corresponding angles is congruent.
How do you identify corresponding angles?
Corresponding angles are the pairs of angles on the same side of the transversal and on corresponding sides of the two other lines. These angles are equal in degree measure when the two lines intersected by the transversal are parallel. It may help to draw the letter “F” (forwards and backwards) in order to help identify corresponding angles.
### What is the formula for corresponding angles?
Corresponding Angles Formula – Trigonometric Angles. Congruent corresponding angles are: Angle of a = Angle of g. Angle of b = Angle of h. Angel of c = Angle of e. Angle of d = Angle of f.
### What is a real life example of a corresponding angle?
Examples in Real Life. Windows have horizontal and vertical grills , which make multiple squares. Each of the vertex makes corresponding angles; A bridge standing on a pillar, where each pillars are connected to each other in such a way that corresponding angles are equal; The design of the railway track where the corresponding angles are kept equal
What is an example of a corresponding angle?
Corresponding Angles When two lines are crossed by another line (which is called the Transversal), the angles in matching corners are called corresponding angles. Example: a and e are corresponding angles. When the two lines are parallel Corresponding Angles are equal. | 802 | 3,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-50 | latest | en | 0.855316 |
https://mathleaks.com/book-series/big-ideas-learning/big-ideas-math-algebra-1-2015/solving-systems-of-linear-equations/quiz | 1,601,480,400,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402127075.68/warc/CC-MAIN-20200930141310-20200930171310-00553.warc.gz | 491,695,082 | 10,527 | #### Quiz
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##### Sections
###### Exercises
Exercise name Free?
Exercises 1 The solution of the system of the linear equations is the point where the graphs of the equations intersect. From the graph, we can tell that this point is (3,1).Let's check this solution by substituting the point into each equation. Note that in order to verify the answer not only the point has to satisfy one of the equations but also the other one. {y=-31x+2y=x−2(I)(II)(I), (II): x=3, y=1{1=?-31⋅3+21=?3−2(I): 3a⋅3=a{1=?-1+21=?3−2(I), (II): Add and subtract terms{1=11=1 Since both equations hold true after substituting the point, (3,1) is a solution of the system of equations.
Exercises 2 The solution of the system of the linear equations is the point where the graphs of the equations intersect. From the given graph, it seems like this point is (-2,-2).We can check this solution by substituting the point into each equation.First equation Let's substitute (-2,-2) into the first equation and simplify. y=21x−1x=-2, y=-2-2=?21(-2)−1a(-b)=-a⋅b-2=?-21⋅2−12a⋅2=a-2=?-1−1Subtract term-2=-2 The equation is true.Second equation Let's now do the same with the second equation. y=4x+6x=-2, y=-2-2=?4⋅(-2)+6a(-b)=-a⋅b-2=?-8+6Add terms-2=-2 The second equation is also true.Conclusion Since both equations hold true after substituting the point, it is the solution of the system of equations.
Exercises 3 The solution of the system of the linear equations is the point where the graphs intersect. From the graph, we can tell that this point is (0,1).Let's check this solution by substituting the point into each equation. Note that in order to verify the answer, the point has to satisfy both of the equations. {y=1y=2x+1(I)(II)(I), (II): x=0, y=1{1=11=?2(0)+1(II): Zero Property of Multiplication{1=11=1 Since both equations hold true after substituting the point, we can confirm that (0,1) is a solution for the system of equations.
Exercises 4 When solving a system of equations using substitution, there are three steps.Isolate a variable in one of the equations. Substitute the expression for that variable into the other equation and solve. Substitute this solution into one of the equations and solve for the value of the other variable. For this exercise, y is already isolated in one equation, so we can skip straight to step 2! {y=x−4-2x+y=18(I)(II)(II): y=x−4{y=x−4-2x+x−4=18(II): Add terms{y=x−4-x−4=18(II): LHS+4=RHS+4{y=x−4-x=22(II): Change signs{y=x−4x=-22 Now, to find the value of y, we need to substitute x=-22 into either one of the equations in the given system. Let's use the first equation. {y=x−4x=-22(I): x=-22{y=-22−4x=-22(I): Subtract term{y=-26x=-22 The solution, or point of intersection, to this system of equations is the point (-22,-26).Checking Our Answer To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct. {y=x−4-2x+y=18(I)(II)(I), (II): x=-22, y=-26{-26=?-22−4-2(-22)+(-26)=?18(II): -a(-b)=a⋅b{-26=?-22−444−26=?18(I), (II): Subtract terms{-26=-2618=18 Because both equations are true statements, we know that our solution is correct.
Exercises 5 In order to use the Substitution Method to solve a system of equations we need to begin by isolating either variable in one of the equations. We can then substitute the expression for that variable into the second equation. We have been given the following system of equations. {2y+x=-4y−x=-5(I)(II) We will start by isolating the y variable in Equation (II). y−x=-5⇔y=-5+x Now we can substitute (II) into (I) and solve for x. {2y+x=-4y=-5+x(I)(II)(I): y=-5+x{2(-5+x)+x=-4y=-5+x(I): Distribute 2{-10+2x+x=-4y=-5+x(I): Add terms{-10+3x=-4y=-5+x(I): LHS+10=RHS+10{3x=6y=-5+x(I): LHS/3=RHS/3{x=2y=-5+x We can now substitute x=2 into Equation (II) to solve for y. {x=2y=-5+x(II): x=2{x=2y=-5+2(II): Add terms{x=2y=-3 The solution to the system is (2,-3).Checking the solution In order to check the solution we need to substitute x by 2 and y by -3 in both equations and verify we end up in an identity. {2y+x=-4y−x=-5(I)(II)(I), (II): x=2, y=-3{2(-3)+2=?-4-3−2=?-5(I): a(-b)=-a⋅b{-6+2=?-4-3−2=?-5(I), (II): Add and subtract terms{-4=-4-5=-5 The solution verifies both equations.
Exercises 6 Using the Substitution Method to solve a system of equations requires us to begin by isolating either variable in one of the equations. We can then substitute the expression for that variable into the second equation. We have been given the following system of equations. {3x−5y=13x+4y=10(I)(II) Before we can substitute one equation into the other, we must alter one of the equations so that it is formatted as x=… or y=…Finding y Since the coefficient of x in (II) is 1, it will only require one step to isolate x. x+4y=10⇔x=10−4y Now we can substitute (II) into (I) and solve for y. {3x−5y=13x=10−4y(I)(II)(I): x=10−4y{3(10−4y)−5y=13x=10−4y(I): Distribute 3{30−12y−5y=13x=10−4y(I): Subtract term{30−17y=13x=10−4y(I): LHS−30=RHS−30{-17y=-17x=10−4y(I): LHS/(-17)=RHS/(-17){y=1x=10−4y The y part of our solution is y=1.Finding x We can now substitute y=1 into either (I) or (II) to solve for x. Let's use (II). {y=1x=10−4y(I)(II)(II): y=1{y=1x=10−4⋅1(II): a⋅1=a{y=1x=10−4(II): Subtract term{y=1x=6 The x part of our solution is x=6, so the solution to the system is (6,1).Checking Our Answer To check our answer, we will substitute the solution we found into both equations. If both equations remain true, our solution is correct. {3x−5y=13x=10−4y(I)(II)x=6, y=1{3(6)−5(1)=13(6)=10−4(1)(I): (II): Multiply{18−5=136=10−4(I): (II): Subtract terms{13=136=6 Both equations are identities, so our solution is correct.
Exercises 7 To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. In this exercise, this means that either the x-terms or the y-terms must cancel each other out. {x+y=4-3x−y=-8(I)(II) We can see that the y-terms will eliminate each other if we add (I) to (II). {x+y=4-3x−y=-8(II): Add (I){x+y=4-3x−y+x+y=-8+4 (II): Solve for x (II): Add and subtract terms{x+y=4-2x=-4(II): LHS/-2=RHS/-2 {x+y=4x=2 Now we can now solve for y by substituting the value of x into either equation and simplifying. Let's use the first equation. {x+y=4x=2(I): x=2{2+y=4x=2(I): LHS−2=RHS−2{y=2x=2 The solution, or intersection point, of the system of equations is (2,2). To check this solution, we will substitute it back into the given system and simplify. If doing so results in true statements for every equation in the system, our solution is correct. {x+y=4-3x−y=-8(I)(II)(I), (II): x=2, y=2{2+2=?4-3(2)−2=?-8(I)(II)(II): (-a)b=-ab{2+2=?4-6−2=?-8(I)(II)(I), (II): Add and subtract terms{4=4-8=-8(I)(II) Because both equations are true statements, we know our solution is correct.
Exercises 8 We will find the solution of the system of equations and then we will check the answer.Finding the solution To use the Elimination Method, either the x- or y-terms in both equations must cancel out when the equations are added. If we multiply the first equation by -2, the coefficients of y will be -6 and 6. Then, when we add the equations, y will cancel out, and we can solve for x. {x+3y=15x+6y=14(I)(II)(I): LHS⋅(-2)=RHS⋅(-2){-2x−6y=-25x+6y=14(I): Add (II){-2x−6y+5x+6y=-2+145x+6y=14(I): Add and subtract terms{3x=125x+6y=14(I): LHS/3=RHS/3{x=45x+6y=14 Now that we've found the value of x, we can substitute it into Equation (II) and solve for y. {x=45x+6y=14(II): x=4{x=45⋅4+6y=14(II): Multiply{x=420+6y=14(II): LHS−20=RHS−20{x=46y=-6(II): LHS/6=RHS/6{x=4y=-1 The solution to the system of equations, which is the point of intersection of the two lines, is (4,-1).Checking the answer To check the answer, we will substitute 4 for x and -1 for y in both equations and simplify. {x+3y=15x+6y=14(I)(II)(I), (II): x=4, y=-1{4+3(-1)=?15⋅4+6(-1)=?14 Simplify LHS (II): Multiply{4+3(-1)=?120+6(-1)=?14(I), (II): a(-b)=-a⋅b{4−3=?120−6=?14(I), (II): Subtract terms {1=114=14 Since we have arrived to two identities, the solution (4,-1) is correct.
Exercises 9 To use the Elimination Method, either the x- or y-terms in both equations must cancel out when the equations are added. If we multiply the first equation by 2 and the second equation by 3, the coefficients of y will be -6 and 6. {2x−3y=-55x+2y=16(I)(II)(I): LHS⋅2=RHS⋅2{4x−6y=-105x+2y=16(II): LHS⋅3=RHS⋅3{4x−6y=-1015x+6y=48 Now when we add the equations, y will cancel out and we can solve for x. {4x−6y=-1015x+6y=48(II): Add (I){4x−6y=-1015x+6y+4x−6y=48+(-10)(II): Add and subtract terms{4x−6y=-1019x=38(II): LHS/19=RHS/19{4x−6y=-10x=2 Substitute x=2 into Equation (I) and solve for y. {4x−6y=-10x=2(I): x=2{4⋅2−6y=-10x=2(I): Multiply{8−6y=-10x=2(I): LHS−8=RHS−8{-6y=-18x=2(I): LHS/(-6)=RHS/(-6){y=3x=2 The solution to the system of equations, the intersection point, is (2,3).Checking the Solution We will check this solution by substituting 2 for x and 3 for y into both equations. {2x−3y=-55x+2y=16(I)(II)(I): (II): x=2, y=3{2⋅2−3⋅3=?-55⋅2+2⋅3=?16(I): (II): Multiply{4−9=?-510+6=?16(I): (II): Add and subtract terms{-5=-516=16 Since our solution satisfies both equations, it is true.
Exercises 10 In this system of equations, at least one of the variables has a coefficient of 1. Therefore, we will approach its solution with the substitution method. When solving a system of equations using substitution, there are three steps.Isolate a variable in one of the equations. Substitute the expression for that variable into the other equation and solve. Substitute this solution into one of the equations and solve for the value of the other variable. Observing the given equations, it looks like it will be simplest to isolate x in the first equation. x−y=1LHS+y=RHS+yx=1+y Now that we've isolated x, we can solve the system by substitution. {x=1+yx−y=6(I)(II)(II): x=1+y{x=1+y1+y−y=6(II): Subtract term{x=1+y1=6 Solving this system of equations resulted in a contradiction; 1 can never be equal to 6. The lines are parallel and do not have a point of intersection. Therefore, the system does not have a solution.
Exercises 11 Let's use the elimination method to solve this system. To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. {6x+2y=162x−y=2(I)(II) In its current state, this won't happen. Therefore, we need to find a common multiple between two variable like-terms in the system. If we multiply (II) by 2, the y-terms will have opposite coefficients. {6x+2y=162(2x−y)=2(2) ⇒ {6x+2y=164x−2y=4 We can see that the y-terms will eliminate each other if we add (I) to (II). {6x+2y=164x−2y=4(II): Add (I){6x+2y=164x−2y+6x+2y=4+16 (II): Solve for x (II): Add and subtract terms{6x+2y=1610x=20(II): LHS/10=RHS/10 {6x+2y=16x=2 Now we can now solve for y by substituting the value of x into either equation and simplifying. {6x+2y=16x=2(I): x=2{6(2)+2y=16x=2 (I): Solve for y (I): Multiply{12+2y=16x=2(I): LHS−12=RHS−12{2y=4x=2(I): LHS/2=RHS/2 {y=2x=2 The solution, or intersection point, of the system of equations is (2,2).
Exercises 12 Since neither equation has a variable with a coefficient of 1, the substitution method may not be the easiest. Instead, let's use the elimination method. To use this method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. {3x−3y=-2-6x+6y=4(I)(II) In its current state, this won't happen. Therefore, we need to find a common multiple between two variable like-terms in the system. If we multiply (I) by 2, both the x-terms the y-terms will have opposite coefficients. {2(3x−3y)=2(-2)-6x+6y)=4 ⇒ {6x−6y=-4-6x+6y=4 We can see that both the x-terms and the y-terms will eliminate each other if we add (I) to (II). {6x−6y=-4-6x+6y=4(II): Add (I){6x−6y=-4-6x+6y+6x−6y=4+(-4)(II): Add and subtract terms{6x−6y=-40=0 Solving this system of equations resulted in an identity; 0 is always equal to itself. This means that the lines are the same and have infinitely many intersection points. Therefore the system has infinitely many solutions.
Exercises 13
Exercises 14
Exercises 15 We are told that the home team scored 6 times. They have collected 26 points from 7-point touchdowns and 3-point. If we let x represent the number of 7-pointers and y represent the number of 3-pointers, we can write the following system of equations. {7x+3y=26x+y=6(I)(II) In order to solve it we will use the Substitution Method. {7x+3y=26x+y=6(I)(II)(II): LHS−x=RHS−x{7x+3y=26y=6−x(I)(II)(I): y=6−x{7x+3(6−x)=26y=6−x(I)(II)(I): Distribute 3{7x+18−3x=26y=6−x(I)(II)(I): Subtract term{4x+18=26y=6−x(I)(II)(I): LHS−18=RHS−18{4x=8y=6−x(I)(II)(I): LHS/4=RHS/4{x=2y=6−x(I)(II)(II): x=2{x=2y=6−2(I)(II)(II): Subtract term{x=2y=4(I)(II) In the context of the word problem, this means that the home team scored two 7-point touchdowns and four 3-point field goals. | 4,880 | 13,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2020-40 | latest | en | 0.860894 |
https://www.coursehero.com/file/5930297/che320-class7edited/ | 1,490,521,500,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189198.71/warc/CC-MAIN-20170322212949-00035-ip-10-233-31-227.ec2.internal.warc.gz | 869,145,229 | 54,494 | che320_class7edited
# che320_class7edited - 7 Residual properties (nonideality)...
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7 Residual properties (nonideality) today . residual properties reference states vapor-liquid equilibrium fugacity of a pure substance announce . 1 st exam (8 am on Fri 18sep2009, 111 Wartik). idterm exam Mon 05oct2009 from 8:15 :15. 114 Midterm exam Mon 05oct2009 from 8:15-11:15. assignments . HW 03 due before class Wed 16sep2009. reading: Ch 03, start Ch 04
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Example. Nonideality of gas volumes For n-butane at 350 K, where its vapor pressure is 9.4573 bar, estimate the non-ideality of the gas volume. answer: -582 cm 3 /mol 115
Residual properties 116
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Residual enthalpy and entropy 117
Residual enthalpy and entropy for SRK, PR 118
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Example. Nonideality of enthalpy and entropy For n-butane at 350 K, where its vapor pressure is 9.4573 bar, estimate the non-ideality of the gas enthalpy and entropy. Find H R and S R using the SRK EOS. answer: H R = -1418 J/mol, S R = -2.96 J/mol-K 119
blank 120
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blank 121
“0” means evaluated at reference state (P 0 , T 0 ). r gases
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## This note was uploaded on 08/22/2010 for the course CH E 320 taught by Professor Matsoukas,themist during the Fall '09 term at Pennsylvania State University, University Park.
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Ask a homework question - tutors are online | 559 | 2,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-13 | longest | en | 0.833579 |
https://storables.com/articles/how-to-read-a-metric-electrical-cord-numbers/ | 1,720,946,282,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00051.warc.gz | 486,229,423 | 60,685 | Home>Articles>How To Read A Metric Electrical Cord Numbers
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# How To Read A Metric Electrical Cord Numbers
Written by: Lily Evans
Looking for articles on how to read metric electrical cord numbers? Check out our comprehensive guide to understand the numbers and symbols on metric electrical cords.
## Introduction
Welcome to the world of metric electrical cord numbers! If you have ever wondered about the mystery behind those numbers printed on electrical cords, you have come to the right place. Understanding these numbers can help you choose the right cord for your specific needs, whether it’s for household appliances, power tools, or industrial applications.
In this article, we will delve into the intricacies of metric electrical cord numbers and break down their components. By the end, you will know how to decipher these numbers, determine cord characteristics, and understand their various uses.
So, let’s embark on this journey of unraveling the secrets behind metric electrical cord numbers!
## Key Takeaways:
• Understanding metric electrical cord numbers is crucial for selecting the right cord. By deciphering the components, you can determine characteristics like current capacity, flexibility, and durability, ensuring safe and efficient power distribution.
• Different applications require specific cord characteristics. Whether for household appliances, power tools, industrial use, outdoor settings, or heavy-duty applications, choosing the appropriate metric electrical cord is essential for safety and reliability.
## Understanding Metric Electrical Cord Numbers
Metric electrical cord numbers consist of a series of digits that provide vital information about the cord’s specifications. These numbers are standardized and follow a specific format, making it easier for consumers and professionals to identify and select the right cord for their needs.
These numbers generally include information about the cord’s gauge, conductor material, insulation type, and voltage rating. By understanding these components, you can make informed decisions when choosing an electrical cord that is safe, efficient, and suitable for your intended application.
It’s important to note that metric electrical cord numbers may vary between different regions or countries, so it’s always recommended to refer to local standards and regulations when purchasing cords.
Now, let’s dive deeper into the components that make up these metric electrical cord numbers.
## Deciphering Cord Number Components
When looking at a metric electrical cord number, you will come across several components that convey important information. Understanding what each component represents will help you identify the cord’s characteristics and capabilities.
1. Gauge: The gauge refers to the thickness or diameter of the cord’s conductors. It is typically represented by a number, such as 16 or 12. The lower the gauge number, the thicker the conductors. Thicker conductors can handle higher current loads, making them suitable for heavy-duty applications.
2. Conductor Material: This component indicates the material used for the cord’s conductors, which carry the electrical current. Common conductor materials include copper and aluminum. Copper is known for its superior conductivity and is often used in high-quality cords.
3. Insulation Type: The insulation of an electrical cord protects the conductors from damage and prevents electrical shocks. It is crucial to choose a cord with insulation suitable for the intended application. Common insulation types include PVC (Polyvinyl Chloride) and rubber.
4. Voltage Rating: The voltage rating specifies the maximum voltage that the cord can safely handle. It is typically expressed in volts (V) or kilovolts (kV). Using a cord with a voltage rating lower than the application’s voltage can lead to electrical hazards and equipment damage.
By paying attention to these cord number components, you can ensure that the cord you choose is compatible with your specific electrical needs and provides reliable performance.
Now that we have deciphered the components, let’s move on to understanding how to determine cord characteristics based on these numbers.
When reading a metric electrical cord number, the first digit represents the number of conductors, the second digit represents the cross-sectional area of the conductors in square millimeters, and the letter indicates the insulation type.
## Determining Cord Characteristics
Once you have deciphered the components of a metric electrical cord number, you can determine various characteristics of the cord. These characteristics play a crucial role in choosing the right cord for your specific application.
1. Current Capacity: The gauge of the cord determines its current-carrying capacity. Thicker conductors can handle higher current loads without overheating. Consult a current capacity chart or table to determine the maximum current that a particular gauge can handle.
2. Flexibility: The flexibility of a cord is influenced by factors such as conductor material and insulation type. Cords with stranded conductors and flexible insulation offer improved flexibility, making them suitable for applications that require frequent bending or movement.
3. Temperature Resistance: Different insulation materials have varying temperature ratings. It is important to consider the temperature conditions in your application and choose a cord that can withstand those temperatures without degrading or melting.
4. Durability: The overall durability of a cord depends on factors such as insulation material, conductor protection, and construction quality. Consider the environment and conditions in which the cord will be used and choose a cord that can withstand the required level of durability.
5. Cord Length: While not directly indicated in the metric electrical cord number, the cord length is an important consideration. Determine the required length based on the distance between the power source and the equipment, ensuring that it offers adequate reach without excessive slack.
By considering these characteristics, you can select a metric electrical cord that meets your specific requirements and ensures optimal performance and safety.
Now, let’s move on to interpreting the various uses of metric electrical cords.
## Interpreting Cord Uses
Metric electrical cords are designed to cater to a wide range of applications. Understanding the different uses associated with these cords will help you select the right one for your specific needs.
1. General Household Appliances: Metric electrical cords with a standard gauge and insulation are commonly used for powering everyday household appliances such as lamps, computers, televisions, and kitchen appliances. These cords are typically flexible and have a voltage rating suitable for residential use.
2. Power Tools: Power tools often require cords with thicker conductors and higher current-carrying capacities to handle the demanding power requirements. Look for metric electrical cords specifically designed for power tools to ensure safe and efficient operation.
3. Industrial Applications: Industrial environments require rugged and durable cords capable of withstanding harsh conditions. These cords often feature heavy-duty construction, thicker conductors, and specialized insulation to meet the demands of industrial machinery and equipment.
4. Outdoor Use: When using electrical equipment outdoors, it is crucial to select cords that are weather-resistant. Look for cords with suitable insulation and coatings that provide protection against moisture, UV rays, and extreme temperatures.
5. Heavy-Duty Applications: Certain applications, such as construction sites or industrial settings, may require cords with exceptional durability and high current-carrying capacities. These cords are designed to handle heavy loads, resist abrasion, and withstand rough handling.
Remember to consider the specific requirements and environmental conditions of your intended application when selecting a metric electrical cord. This will ensure that the cord is up to the task and can provide safe and reliable power distribution.
Now that we have covered the uses of metric electrical cords, let’s summarize our findings.
## Conclusion
Understanding metric electrical cord numbers is essential for selecting the right cord for your specific needs. By deciphering the components and understanding their meanings, you can determine the cord’s characteristics, such as current capacity, flexibility, temperature resistance, and durability.
When interpreting cord uses, consider the specific application, whether it is for general household appliances, power tools, industrial settings, outdoor use, or heavy-duty applications. Each use may require different characteristics, such as voltage rating, insulation type, and construction quality.
By choosing the appropriate metric electrical cord, you can ensure efficient and safe power distribution, preventing electrical hazards and equipment damage. It is crucial to consult local standards and regulations to ensure compliance and safety.
Remember to consider factors like cord length, conductor material, insulation type, and overall quality when making your selection. Additionally, if you have any doubts or need assistance, don’t hesitate to seek advice from professionals or knowledgeable sources.
With your newfound knowledge about metric electrical cord numbers, you can confidently navigate the world of electrical cords and make informed decisions that align with your specific requirements.
Now, armed with this information, go forth and choose the perfect metric electrical cord for your needs!
What do the numbers on a metric electrical cord mean?
The numbers on a metric electrical cord typically indicate the cord’s size, which includes the cross-sectional area of the conductor and the overall diameter of the cord. Understanding these numbers is essential for selecting the right cord for your specific electrical needs.
How can I determine the appropriate metric electrical cord for my project?
To determine the appropriate metric electrical cord for your project, you should consider the voltage and current requirements, as well as the environmental conditions the cord will be exposed to. Additionally, understanding the metric measurements and their corresponding ampacity ratings will help you make an informed decision.
Why is it important to read and understand metric electrical cord numbers?
It is important to read and understand metric electrical cord numbers to ensure the safety and efficiency of your electrical system. Using the wrong cord size or type can lead to overheating, voltage drop, and potential hazards. By understanding the numbers, you can make informed choices that comply with electrical codes and regulations.
Can I use a metric electrical cord with different numbers than the original cord?
It is not recommended to use a metric electrical cord with different numbers than the original cord specified for your equipment. The numbers indicate the cord’s capacity to handle specific electrical loads, and using a cord with different numbers may result in inadequate power delivery or potential safety risks.
One common misconception is assuming that a higher number always means a better or more powerful cord. In reality, the numbers on a metric electrical cord are specific to its intended use and should be carefully matched to the electrical requirements of the equipment or application. | 2,001 | 11,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-30 | latest | en | 0.856859 |
https://www.physicsforums.com/threads/limits-and-lhopital-rule.101441/ | 1,623,703,943,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00413.warc.gz | 840,301,760 | 15,294 | # Limits and L'Hopital rule
I have to solve this:
$$\lim_{\substack{s\rightarrow 0^+}} s^4 (\frac{1}{2} ln (s) - \frac{1}{8})$$
Here is what I did so far:
$$\lim_{\substack{s\rightarrow 0^+}} \frac{s^4}{\frac{1}{\frac{1}{2} ln (s) - \frac{1}{8}} =$$
$$= \lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{\frac{-\frac{1}{2s}}{(\frac{1}{2} ln (s) - \frac{1}{8})^2}} =$$
$$= \lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}} =$$
$$= \lim_{\substack{s\rightarrow 0^+}} \frac{12s^2}{\frac{2 (\frac{1}{2} ln (s) - \frac{1}{8})^2 + 2 (\frac{1}{2} ln (s) - \frac{1}{8})} {[2s (\frac{1}{2} ln (s) - \frac{1}{8})^2]^2}} = 0$$
Is this OK? If not, can someone help me please?
Last edited:
I think you didn't get the product rule quite right in your second use of L'Hospital's.
You may find it simpler if you look at (1/2)Ln(s) to be the Ln (s^1/2)
(It MAY reduce the problem to only one use of L'Hospital's) :)
Last edited:
shmoe
Homework Helper
Did you stop to check if the conditions of l'hopital were satisfied before you applied it a second time? What is
$$\lim_{\substack{s\rightarrow 0^+}}-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}$$
You might want to try going back to the start and try a different rearrangement to get it into a form you can apply l'hopital to.
You mean that:
$$\lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}} = 0$$
And that the problem ends there?
Last edited:
HallsofIvy
Homework Helper
No, the problem doesn't end until you know what the limit is! (Or show that there is no limit.) Do what Shmoe said, "You might want to try going back to the start and try a different rearrangement to get it into a form you can apply l'hopital to."
Sorry, I don't understand what you mean. I arranged the function in the first step to get a "0/0" indetermination and applied L'Hopital to that.
Can you show me what you would do please?
shmoe
Homework Helper
I just want to point out that at this stage:
$$\lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}}$$
this limit is the same as:
$$\lim_{\substack{s\rightarrow 0^+}} -8s^4(\frac{1}{2} ln (s) - \frac{1}{8})^2$$
Which is more complicated than what you started with- the polynomial has the same power, and the icky term with the log has a higher power making it even ickier. This will most likely be harder to deal with than the original limit so it's a big clue you should go back to the drawing board and modify your approach.
There's more than one way to arrange what you started with to an indeterminate form you can apply l'hopital to!
Last edited: | 936 | 2,688 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-25 | latest | en | 0.797356 |
https://techcommunity.microsoft.com/t5/excel/excel-formula-for-lookup/td-p/3916212 | 1,719,099,654,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00803.warc.gz | 489,738,865 | 54,507 | # Excel formula for lookup
Copper Contributor
# Excel formula for lookup
We have two sets of data to compare. In one set of data, cell A1 has numbers separated by commas, like this:
13010982231, 13010982232, 13011011681, 13011011682
Cell A2 has the name of the person, example, James, and James is considered to be associated to those numbers.
DATA SET1
Number Name
1301086213 Jim 1301088551 Joe 1301058852 Melinda 1301032770 Melissa 13010750451 Billy 1301078972 Johnny 1301096578 Abigal 1301055408, 1301055619, 1301055901 Chris
In the other data set, we just have one number in a cell and we want to check and see if that number in the cell can be found in the list of numbers in Column A from the other data set1, and if it is found, tell us the name of the person that number is associated to in that data set and fill that name into the name column in DATA SET2.
DATA SET2
Number NAME 13010914141 13010914142 13010821991 13010821911 13011037061 13011037062 13010979211 13011026881 13011032761 13011032762 13011032763 13011032764 13011032765
We used to be able to do this using VLOOKUP, because there was only one number in the cell, but now that the cell contains multiple numbers, it doesn't work as expected. We are able to lookup some of the information by using a vlookup formula similar to this, =VLOOKUP("*"&G2&"*",DATA,2,FALSE) but the results do not always populate. | 403 | 1,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.816914 |
https://tackk.com/mtzefy | 1,503,374,380,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109893.47/warc/CC-MAIN-20170822031111-20170822051111-00347.warc.gz | 849,664,843 | 7,272 | # Newton's laws of Motion
### By:Carlie #NewtonsLaws#Kettner
Newton's First Law of Motion
Newton's law of Inertia
Newton's First Law of motion states that an object in motion tend to stay in motion and an object at rest tends to stay at rest. In this example the biker was moving fast ion his bike and ran into a wall, causing the bike to stop but the biker to keep moving forward.
Newton's Second Law of Motion
Newton's Second Law of Motion states that acceleration = mass x speed. This example shows a person pushing a car weighing 1000 kg at an unknown rate of force equaling the acceleration of 0.05 m/s, in which case you would need to find the force being applied to the car.
Newton's Third Law of Motion
Newton's Third law of Motion states that an object acted upon with force will have an equal and opposite reaction. In this example the stick figure is standing on the edge of the boat, the reaction would be, if the boat was sturdy enough, to remain floating, or tipping over along with the stick figure. | 236 | 1,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-34 | latest | en | 0.961502 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-11-section-11-3-limits-and-continuity-exercise-set-page-1162/65 | 1,726,431,174,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00005.warc.gz | 721,241,240 | 12,523 | ## Precalculus (6th Edition) Blitzer
The angle between $v=\text{5}i+2j\text{ and }w=-3i+j$ is ${{139.8}^{\circ }}$.
Consider the vectors, $v=\text{5}i+2j\text{ and }w=-3i+j$. The angle between $v=\text{5}i+2j\text{ and }w=-3i+j$ is, $\theta ={{\cos }^{-1}}\frac{\left( 5i+2j \right)\times \left( -3i+j \right)}{\left\| v \right\|\times \left\| w \right\|}$ To find the value of $\left\| v \right\|\text{ and }\left\| w \right\|$, \begin{align} & \left\| v \right\|=\sqrt{{{5}^{2}}+{{2}^{2}}} \\ & =\sqrt{25+4} \\ & =\sqrt{29} \end{align} And \begin{align} & \left\| w \right\|=\sqrt{{{\left( -3 \right)}^{2}}+{{1}^{2}}} \\ & =\sqrt{9+1} \\ & =\sqrt{10} \end{align} Substitute the values of $\left\| v \right\|\text{ and }\left\| w \right\|$ in $\theta ={{\cos }^{-1}}\frac{\left( 5i+2j \right)\times \left( -3i+j \right)}{\left\| v \right\|\times \left\| w \right\|}$ and solve for $\theta$, \begin{align} & \theta ={{\cos }^{-1}}\frac{\left( 5i+2j \right)\times \left( -3i+j \right)}{\left( \sqrt{29} \right)\left( \sqrt{10} \right)} \\ & ={{\cos }^{-1}}\frac{-15+2}{\left( \sqrt{29} \right)\left( \sqrt{10} \right)} \\ & ={{\cos }^{-1}}\frac{-13}{\sqrt{290}}\approx {{139.8}^{\circ }} \end{align} Hence, the angle between $v=\text{5}i+2j\text{ and }w=-3i+j$ is ${{139.8}^{\circ }}$. | 561 | 1,285 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-38 | latest | en | 0.418117 |
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# Logical and shift micro operations
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### Logical and shift micro operations
1. 1. PRESENTED BY:<br /> ASHISH SIKKA<br />LOGICAL AND SHIFT MICROOPERATION<br />
2. 2. Logic microoperation specify binary operation for strings of bit stored in registers.<br />These operation consider each bit of the register separately and treat them as binary variables.For example,<br /> P:R1 R1 R2<br /> 1010 Content of R1<br /> 1100 Content of R2<br /> 0110 Content of R1 after P=1 <br />WHAT IS LOGIC MICROOPERATION<br />
3. 3. LIST OF logic microoperation<br />
4. 4. Sixteen logic microoperation<br />
5. 5. Hardware implementation <br />
6. 6. <ul><li>SELECTIVE SET:</li></ul> it sets the bit’s to 1 in register A where there are 1,s in register B. example<br /> 1010 A before<br /> 1100 B(logic operant)<br /> 1110 A after<br /><ul><li>SELECTIVE COMPLEMENT : </li></ul> it complements bits in A where there are corresponding 1’s in B.example<br /> 1010 A before<br /> 1100 B<br /> 0110 A after<br />it can be seen selective complement can be done by Exclusive –OR<br />Some other function<br />
7. 7. SELECTIVE CLEAR:<br /> it clear the bit to 0 in A where there are corresponding 1’s in B.example<br /> 1010 A before<br /> 1100 B<br /> 0010 A after<br />(it can be obtained by microoperation AB’)<br />MASKING:<br /> it is similar to selective clear except that the bit of A is cleared where there corresponding 0’s.<br /> 1010 A before<br /> 1100 B<br /> 1000 A after<br />
8. 8. INSERT :<br /> it inserts a new value into a group of bits.<br /> This is done by first masking and then Oring with the value.Example<br /> 0110 1010 A before<br /> 0000 1111 B<br /> 0000 1010 A after<br /> then insert a new value<br /> 0000 1010 A before<br /> 1001 0000 B(insert)<br /> 1001 1010 A after<br />
9. 9. Shift microoperation are used for serial transfer of data.The content of the register can be shifted to left or the right.At the same time thet the bits are shifted the the first flip flop receive its binary information from the serial input.The information transferred through the serial input determines the type of shift.<br />There are three types of shift:<br />Logical shift<br />Circular shift<br />Arithmetic shift <br />SHIFT MICROOPERATION<br />
10. 10. A logical shift is one that transfer 0 through the serial input.The bit transferred to the end position through the serial input is assumed to be zero.<br />Example:<br /> R1 shl R1 (1 bit shift to the left)<br /> R2 shr R2(1 bit shift to the right)<br />Logical shift<br />
11. 11. The circular shift(also known as rotate operation) circulates the bits of the register around the ends without the loss of information.<br />This is accomplished by the connecting the serial output of the register to the serial input.<br />Example:<br /> R1 cil R1(shifts left)<br /> R2 cir R2(shifts right)<br />Circular shift<br />
12. 12. An arithmetic shift is a microoperation that shifts signed binary number to the left or right.<br />An arithmetic shift left multiplies a signed binary no. by 2 and shift right divides by 2.<br />The signed bit remains unchanged whether it is divided or multiplied by 2.<br />arithmetic shift<br />
13. 13. The arithmetic shift right leaves the sign bit unchanged and shift the no.(including the sign bit) to the right the bit Rn-1 remain unchanged and R0 is lost.<br />The arithmetic shift left insert a 0 into R0 and shifts all the other bits to the left.The initial bit of Rn-1 is lost and replaced by the bit from Rn-2.A sign reversal occurs if the bit in Rn-1 changes in the value after shift.This happens if the multiplication by 2 causes an overflow.<br />An over-flow flip-flop Vs can be used to detect an arithmetic shift left overflow.<br /> Vs=Rn-1 Rn-2<br />If Vs=0,there is no over flow,if Vs=1 there is overflow and a sign reversal takes place.Vs must be transferred into the over flow with the same clock pulse that shifts the register.<br />
14. 14. Hardware implementation<br />
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https://electronics2electrical.com/6026/ | 1,563,606,653,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526489.6/warc/CC-MAIN-20190720070937-20190720092937-00392.warc.gz | 390,867,789 | 22,593 | # On a ladder resting on smooth ground and leaning against vertical wall, the force of friction will be?
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On a ladder resting on smooth ground and leaning against vertical wall, the force of friction will be?
On a ladder resting on smooth ground and leaning against vertical wall, the force of friction will be upwards at its upper end.
## Related questions
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In a d.c. machine the force on a conductor placed on the periphery of the conductor is (a) directly proportional to parallel path (b) inversely proportional to the parallel path (c) inversely proportional to the length of the conductor (d) none of these | 1,084 | 4,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-30 | longest | en | 0.941431 |
https://www.man7.org/linux/man-pages/man3/asinl.3.html | 1,702,061,595,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100769.54/warc/CC-MAIN-20231208180539-20231208210539-00803.warc.gz | 954,057,498 | 3,542 | asin(3) — Linux manual page
```asin(3) Library Functions Manual asin(3)
```
NAME top
``` asin, asinf, asinl - arc sine function
```
LIBRARY top
``` Math library (libm, -lm)
```
SYNOPSIS top
``` #include <math.h>
double asin(double x);
float asinf(float x);
long double asinl(long double x);
Feature Test Macro Requirements for glibc (see
feature_test_macros(7)):
asinf(), asinl():
_ISOC99_SOURCE || _POSIX_C_SOURCE >= 200112L
|| /* Since glibc 2.19: */ _DEFAULT_SOURCE
|| /* glibc <= 2.19: */ _BSD_SOURCE || _SVID_SOURCE
```
DESCRIPTION top
``` These functions calculate the principal value of the arc sine of
x; that is the value whose sine is x.
```
RETURN VALUE top
``` On success, these functions return the principal value of the arc
sine of x in radians; the return value is in the range
[-pi/2, pi/2].
If x is a NaN, a NaN is returned.
If x is +0 (-0), +0 (-0) is returned.
If x is outside the range [-1, 1], a domain error occurs, and a
NaN is returned.
```
ERRORS top
``` See math_error(7) for information on how to determine whether an
error has occurred when calling these functions.
The following errors can occur:
Domain error: x is outside the range [-1, 1]
errno is set to EDOM. An invalid floating-point exception
(FE_INVALID) is raised.
```
ATTRIBUTES top
``` For an explanation of the terms used in this section, see
attributes(7).
┌──────────────────────────────────────┬───────────────┬─────────┐
│Interface │ Attribute │ Value │
├──────────────────────────────────────┼───────────────┼─────────┤
│asin(), asinf(), asinl() │ Thread safety │ MT-Safe │
└──────────────────────────────────────┴───────────────┴─────────┘
```
STANDARDS top
``` C11, POSIX.1-2008.
```
HISTORY top
``` C99, POSIX.1-2001.
The variant returning double also conforms to SVr4, 4.3BSD, C89.
```
``` acos(3), atan(3), atan2(3), casin(3), cos(3), sin(3), tan(3) | 584 | 2,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-50 | latest | en | 0.443431 |
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